目录
- Bzoj 近期题目题解
- 1000: A+B Problem (模拟)
- 1008: [HNOI2008]越狱 (容斥)
- 1012: [JSOI2008]最大数maxnumber (线段树)
- 1034: [ZJOI2008]泡泡堂BNB (贪心)
- 1036: [ZJOI2008]树的统计Count (树链剖分)
- 1051: [HAOI2006]受欢迎的牛 (tarjan)
- 1059: [ZJOI2007]矩阵游戏 (二分图匹配)
- 1355: [Baltic2009]Radio Transmission (KMP)
- 1192: [HNOI2006]鬼谷子的钱袋 (二进制)
- 1407: [Noi2002]Savage (exgcd,数论)
- 1477: 青蛙的约会 (exgcd,数论)
- 1798: [Ahoi2009]Seq 维护序列seq (线段树)
- 2208: [Jsoi2010]连通数 (bitset)
- 2875: [Noi2012]随机数生成器 (矩阵快速幂)
- 3224: Tyvj 1728 普通平衡树 (splay)
- 3555: [Ctsc2014]企鹅QQ (hash)
- 4034: [HAOI2015]树上操作 (树链剖分)
Bzoj 近期题目题解
Bzoj的简单题.我做过的题目.
1000: A+B Problem (模拟)
copy下面即可
#include <iostream>
using namespace std;
int main()
{
int a,b;
cin >> a >> b;
cout << a+b << endl;
return 0;
}
1008: [HNOI2008]越狱 (容斥)
容斥一下就好了.注意有负数
// luogu-judger-enable-o2
#include <iostream>
#include <cstdio>
using namespace std;
long long fast_power(long long a,long long b,long long mod) {
long long ans = 1;
for(long long now = a;b;b >>= 1,now = now * now % mod) {
if(b & 1) ans = ans * now % mod;
}
return ans % mod;
}
int main() {
long long n,m;
scanf("%lld%lld",&m,&n);
printf("%lld",(fast_power(m,n,100003) - m * fast_power(m - 1,n - 1,100003) % 100003 + 100003) % 100003);
return 0;
}
1012: [JSOI2008]最大数maxnumber (线段树)
线段树提前开好空间,编号依次往后挪就好了.
#include <iostream>
#include <cstdio>
#define ll long long
const ll maxN = 200000 + 7;
char s[3];
struct Node {
ll l,r,maxx;
}tree[maxN << 2];
ll max(ll x,ll y) {
return x > y ? x : y;
}
void updata(ll now) {
tree[now].maxx = max(tree[now << 1].maxx,tree[now << 1 | 1].maxx);
return ;
}
inline ll read() {
ll x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
void build(ll l,ll r,ll now) {
tree[now].l = l;tree[now].r = r;
if(l == r) {tree[now].maxx = -0x7fffffff;return;}
ll mid = (l + r) >> 1;
build(l,mid,now << 1);
build(mid + 1,r,now << 1 | 1);
updata(now);
return;
}
void change(ll pos,ll val,ll now) {
if(tree[now].l == tree[now].r) {
tree[now].maxx = val;
return;
}
ll mid = (tree[now].l + tree[now].r) >> 1;
if(pos <= mid) change(pos,val,now << 1);
else change(pos,val,now << 1 | 1);
updata(now);
return ;
}
ll query(ll l,ll r,ll now) {
if(tree[now].l >= l && tree[now].r <= r) {
return tree[now].maxx;
}
ll maxx = -0x7fffffff;
ll mid = (tree[now].l + tree[now].r) >> 1;
if(l <= mid) maxx = max(maxx,query(l,r,now << 1));
if(r > mid) maxx = max(maxx,query(l,r,now << 1 | 1));
return maxx;
}
int main() {
ll n,P;
scanf("%lld%lld",&n,&P);
char c;
ll pos = 0,t = 0;
build(1,n,1);
for(ll i = 1,x;i <= n;++ i) {
scanf("%s%lld",&s,&x);
if(s[0] == 'A') {
ll tmp = ( (x % P) + t % P ) % P;
++ pos;
change(pos,tmp,1);
}
else {
t = query(pos - x + 1,pos,1);
printf("%lld
",t);
}
}
return 0;
}
1034: [ZJOI2008]泡泡堂BNB (贪心)
贪心即可.
#include <iostream>
#include <cstdio>
#include <algorithm>
#define X 100007
int n;
int work(int a[],int b[])
{
int la = 1,lb = 1,ra = n,rb = n,ans = 0;
while(la <= ra && lb <= rb)
{
if(a[la] > b[lb])++ la,++ lb,ans += 2;
else if(a[ra] > b[rb])-- ra,-- rb,ans += 2;
else{
if(a[la] == b[rb])ans ++;
++ la,-- rb;
}
}
return ans;
}
int a[X],b[X];
int main(){
scanf("%d",&n);
for(int i = 1;i <= n;++ i)
scanf("%d",&a[i]);
for(int i = 1;i <= n;++ i)
scanf("%d",&b[i]);
std::sort(a + 1,a + n + 1);
std::sort(b + 1,b + n + 1);
int ans = work(a,b);
printf("%d %d",ans,2 * n - work(b,a));
}
1036: [ZJOI2008]树的统计Count (树链剖分)
树链剖分模板题.
// luogu-judger-enable-o2
#include <iostream>
#include <cstdio>
const int maxN = 3e5 + 7;
using namespace std;
int pos[maxN],top[maxN],size[maxN],dep[maxN],fa[maxN];
int head[maxN],a[maxN],n,m,cut,maxsize;
struct Node{
int l,r,sum,max;
}tree[maxN << 2];
struct edge{
int v,nex;
}e[maxN << 1];
inline int read() {
int x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
void add(int u,int v) {
e[++ cut].nex = head[u];
head[u] = cut;
e[cut].v = v;
}
void build(int l,int r,int now) {
tree[now].l = l;tree[now].r = r;
if(l == r) {
tree[now].max = tree[now].sum = 0;
return;
}
int mid = (l + r) >> 1;
build(l,mid,now << 1);
build(mid + 1,r,now << 1 | 1);
return;
}
void dfs1(int u) {
size[u] = 1;
for(int i = head[u];i;i = e[i].nex ) {
int v = e[i].v;
if(v == fa[u]) continue;
fa[v] = u;dep[v] = dep[u] + 1;
dfs1(v);
size[u] += size[v];
}
return;
}
void dfs2(int u,int top1) {
int k = 0;
maxsize ++;
pos[u] = maxsize;
top[u] = top1;
for(int i = head[u];i;i = e[i].nex) {
int v = e[i].v;
if(dep[v] > dep[u] && size[k] < size[v]) k = v;
}
if( !k )return;
dfs2(k,top1);
for(int i = head[u];i;i = e[i].nex) {
int v = e[i].v;
if(v != k && dep[v] > dep[u]) {
dfs2(v,v);
}
}
return;
}
void change(int now,int pos,int val) {
if(tree[now].l == tree[now].r) {
tree[now].sum = tree[now].max = val;
return;
}
int mid = (tree[now].l + tree[now].r) >> 1;
if(pos <= mid) change(now << 1,pos,val);
if(pos > mid) change(now << 1 | 1,pos,val);
tree[now].sum = tree[now << 1].sum + tree[now << 1 | 1].sum;
tree[now].max = max(tree[now << 1].max,tree[now << 1 | 1].max);
return;
}
int querysum(int now,int l,int r) {
if(tree[now].l >= l && tree[now].r <= r) return tree[now].sum;
int mid = (tree[now].l + tree[now].r) >> 1;
int Ans = 0;
if(l <= mid) Ans += querysum(now << 1,l,r);
if(r > mid) Ans += querysum(now << 1 | 1,l,r);
return Ans;
}
int querymax(int now,int l,int r) {
if(tree[now].l >= l && tree[now].r <= r) return tree[now].max;
int mid = (tree[now].l + tree[now].r) >> 1,total = -1e9;
if(l <= mid) total = max(total,querymax(now << 1,l,r));
if(r > mid) total = max(total,querymax(now << 1 | 1,l,r));
return total;
}
int slovemax(int x,int y) {
int ans = -1e9;
while(top[x] != top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x,y);
ans = max(ans,querymax(1,pos[top[x]],pos[x]));
x = fa[top[x]];
}
if(pos[x] > pos[y]) swap(x,y);
ans = max(ans,querymax(1,pos[x],pos[y]));
return ans;
}
int slovesum(int x,int y) {
int ans = 0;
while(top[x] != top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x,y);
ans += querysum(1,pos[top[x]],pos[x]);
x = fa[top[x]];
}
if(pos[x] > pos[y]) swap(x,y);
ans += querysum(1,pos[x],pos[y]);
return ans;
}
int main() {
// freopen("1.in","r",stdin);
// freopen("1.out","w",stdout);
int x,y;
n = read();
for(int i = 1;i < n;++ i) {
x = read();y = read();
add(x,y);add(y,x);
}
for(int i = 1;i <= n;++ i)
a[i] = read();
dfs1(1);dfs2(1,1);
build(1,maxsize,1);
for(int i = 1;i<= n;++ i)
change(1,pos[i],a[i]);
m = read();char ch[11];
while(m -- ) {
scanf("%s",ch);
scanf("%d%d",&x,&y);
if(ch[0] == 'C') change(1,pos[x],y);
else {
if(ch[1] == 'M') printf("%d
",slovemax(x,y));
else printf("%d
",slovesum(x,y));
}
}
}
1051: [HAOI2006]受欢迎的牛 (tarjan)
tarjan 缩点,被所有牛指向的就是受欢迎的牛们
#include <iostream>
#include <cstdio>
#define max(a,b) a > b ? a : b
#define min(a,b) a > b ? b : a
const int maxN = 10000 + 7;
const int maxM = 50000 + 7;
int dfn[maxN],low[maxN];
int cnt;//时间戳
int belong[maxN],num;//锁点
int ska[maxN],top; //栈
int cu[maxN];
int Size[maxN];
inline int read() {
int x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
struct Node{
int v,nex;
}Map[maxM];
int head[maxN],G;
inline void add_Node(int u,int v) {
Map[++ G] = (Node) {v,head[u]};
head[u] = G;
}
void dfs(int u) {
dfn[u] = low[u] = ++ cnt;
ska[++ top] = u;
for(int i = head[u];i;i = Map[i].nex) {
int v = Map[i].v;
if(!dfn[v]) {
dfs(v);
low[u] = min(low[u],low[v]);
}else if(!belong[v]) low[u] = min(low[u],dfn[v]);
}
if(dfn[u] == low[u]) {
num ++;
while(ska[top] != u) belong[ska[top --]] = num,Size[num] ++;
belong[ska[top --]] = num;
Size[num] ++;
}
}
int main() {
int n,m;
scanf("%d%d",&n,&m);
int u,v;
for(int i = 1;i <= m;++ i) {
scanf("%d%d",&u,&v);
add_Node(u,v);
}
for(int i = 1;i <= n;++ i)
if(!dfn[i]) dfs(i);
for(int i = 1;i <= n;++ i) {
int bel = belong[i];
for(int j = head[i];j;j = Map[j].nex) {
int v = Map[j].v;
int belv = belong[v];
if(belv != bel) cu[bel] ++;
}
}
int tmp = 0,id;
for(int i = 1;i <= num;++ i) {
if(!cu[i]) tmp ++,id = i;
}
if(tmp > 1) return puts("0"),0;
printf("%d",Size[id]);
return 0;
}
1059: [ZJOI2007]矩阵游戏 (二分图匹配)
二分图匹配,将行,列中的黑白子连边.
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
const int maxN = 2000 + 7;
const int maxM = 2000000 + 7;
const int inf = 0x7fffffff;
using namespace std;
int f[507][507];
struct Node {
int v,nex,w;
}map[maxM];
int dep[maxN],head[maxN],S,t,num,n,maxsize;
int id[maxN];
inline int read() {
int x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
void init() {
maxsize = 0;
num = -1;
memset(head,-1,sizeof(head));
return;
}
void add_Node(int u,int v,int w) {
map[++ num].v = v;
map[num].w = w;
map[num].nex = head[u];
head[u] = num;
return ;
}
void add(int u,int v,int w) {
add_Node(u,v,w);
add_Node(v,u,0);
}
bool bfs() {
queue<int>q;
memset(dep,0,sizeof(dep));
while(!q.empty()) q.pop();
q.push(S);
dep[S] = 1;
do{
int u = q.front();
q.pop();
for(int i = head[u];i != -1;i = map[i].nex) {
int v = map[i].v;
if( !dep[v] && map[i].w) {
q.push(v);
dep[v] = dep[u] + 1;
}
}
}while(!q.empty());
if(dep[t]) return true;
return false;
}
int min(int a,int b) {
return a > b ? b : a;
}
int dfs(int now,int dist) {
if(now == t) return dist;
for(int i = head[now];i != -1;i = map[i].nex) {
int v = map[i].v;
if(dep[v] == dep[now] + 1 && map[i].w) {
int di = dfs(v,min(dist,map[i].w));
if(di) {
map[i].w -= di;
map[i ^ 1].w += di;
return di;
}
}
}
return 0;
}
void Dinic() {
int Ans = 0;
while(bfs()) {
while(int d = dfs(S,inf)) {
Ans += d;
}
}
if(Ans == n) puts("Yes");
else puts("No");
return;
}
int main() {
int T;
T = read();
while(T --) {
init();
n = read();
for(int i = 1;i <= n;++ i)
for(int j = 1;j <= n;++ j)
f[i][j] = read();
S = ++ maxsize;
for(int i = 1;i <= n;++ i)
id[i] = ++ maxsize;
for(int i = 1;i <= n;++ i)
id[n + i] = ++ maxsize;
t = ++ maxsize;
for(int i = 1;i <= n;++ i)
add(S,id[i],1);
for(int i = 1;i <= n;++ i)
add(id[i + n],t,1);
for(int i = 1;i <= n;++ i)
for(int j = 1;j <= n;++ j)
if(f[j][i]) add(id[j],id[i + n],1);
for(int i = 1;i <= n;++ i)
for(int j = 1;j <= n;++ j)
if(f[i][j]) add(id[i],id[j + n],1);
Dinic();
}
}
1355: [Baltic2009]Radio Transmission (KMP)
(n - nex[n])是最小循环节.
#include <iostream>
#include <cstdio>
using namespace std;
const int maxN = 1000000 + 7;
int len2;
char s2[maxN];
int nex[maxN];
void get_nex() {
int p = 0;nex[1] = 0;
for(int i = 2;i <= len2;++ i) {
while(s2[i] != s2[p + 1] && p > 0) p = nex[p];
if(s2[i] == s2[p + 1]) p ++;
nex[i] = p;
}
printf("%d",len2 - nex[len2]);
return ;
}
int main() {
scanf("%d",&len2);
cin >> (s2 + 1);
get_nex();
return 0;
}
1192: [HNOI2006]鬼谷子的钱袋 (二进制)
二进制即可
#include <iostream>
#include <cstdio>
int main(){
int n,tot = 0;
scanf("%d",&n);
while(n){
n >>= 1;
tot ++;
}
printf("%d",tot);
}
1407: [Noi2002]Savage (exgcd,数论)
枚举(m),然后(exgcd)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct shadow
{
int c,p,l;
}s[1000];
int x,y;
int gcd(int a,int b) {
return !b ? a : gcd(b,a % b);
}
void exgcd(int a,int b,int &x,int &y) {
if(!b) {
x = 1;
y = 0;
return;
}
exgcd(b,a % b,x,y);
int tmp = x;
x = y;
y = tmp - a / b * x;
}
int n;
bool work(int m)
{
int a,b,c,t,x,y;
for(int i = 1;i <= n;i ++)
{
for(int j = i + 1;j <= n;j ++)
{
a = s[i].p - s[j].p;
c = s[j].c - s[i].c;
b = m;
t = gcd(a,b);
if(c % t == 0)
{
a /= t;b /= t;c /= t;
exgcd(a,b,x,y);
b = abs(b);
x = ( (c * x) % b + b ) % b;
if(!x)
x += b;
if(x <= min(s[i].l,s[j].l))
return 0;
}
}
}
return 1;
}
int main()
{
cin >> n;
int maxx = 0;
for(int i = 1;i <= n;i ++)
{
cin >> s[i].c >> s[i].p >> s[i].l;
maxx = max(maxx,s[i].c);
}
for(int i = maxx;;i ++)
if(work(i))
{
cout << i << endl;
return 0;
}
return 0;
}
1477: 青蛙的约会 (exgcd,数论)
(exgcd)
#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
inline ll read() {
ll x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c- '0';c = getchar();}
return x * f;
}
ll gcd(ll a,ll b) {
return !b ? a : gcd(b ,a % b);
}
ll exgcd(ll a,ll b,ll &x,ll &y) {
if( !b ) {
x = 1;
y = 0;
return a;
}
ll g = exgcd(b,a % b,x,y);
ll tmp = x;
x = y;
y = tmp - (a / b) * y;
return g;
}
ll x,y,x1,y1,k1,L;
int main() {
x = read();y = read();x1 = read();y1 = read();L = read();
if(x1 < y1) {
swap(x1,y1);
swap(x,y);
}
ll tmpx,tmpy,g;
if((y - x) % gcd((x1 - y1),L) != 0) {
printf("Impossible");
return 0;
}
g = exgcd(x1 - y1,L,tmpx,tmpy);
tmpx = tmpx * (y - x) / g;
L = L / g;
tmpx = (tmpx % L + L) % L;
printf("%lld",tmpx);
return 0;
}
1798: [Ahoi2009]Seq 维护序列seq (线段树)
线段树模板题
// luogu-judger-enable-o2
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#define ll long long
const ll maxN = 1e5 + 7;
ll n,m,P;
ll a[maxN];
inline ll read() {
ll x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
struct Node{
struct emmmm{
ll l,r,sum,tag,lazy;
}tree[maxN << 2];
inline void updata(ll now) {
tree[now].sum = (tree[now << 1].sum + tree[now << 1 | 1].sum) % P;
return;
}
void build(ll now,ll l,ll r) {
tree[now].l = l;tree[now].r = r;
tree[now].lazy = 0;
tree[now].tag = 1;
if(l == r) {
tree[now].sum = a[l];
return;
}
ll mid = (l + r) >> 1;
build(now << 1,l,mid);
build(now << 1 | 1,mid + 1,r);
updata(now);
}
void pushdown(ll now) {
ll tag = tree[now].tag;
ll lazy = tree[now].lazy;
tree[now << 1].lazy = (tree[now << 1].lazy * tag + lazy) % P;
tree[now << 1 | 1].lazy = (tree[now << 1 | 1].lazy * tag + lazy) % P;
tree[now << 1].tag = (tree[now << 1].tag * tag) % P;
tree[now << 1 | 1].tag = (tree[now << 1 | 1].tag * tag) % P;
tree[now << 1].sum = (tree[now << 1].sum * tag + (tree[now << 1].r - tree[now << 1].l + 1) * lazy) % P;
tree[now << 1 | 1].sum = (tree[now << 1 | 1].sum * tag + (tree[now << 1 | 1].r - tree[now << 1 | 1].l + 1) * lazy) % P;
tree[now].tag = 1;
tree[now].lazy = 0;
return;
}
void muladd(ll l,ll r,ll k,ll now) {
if(tree[now].l >= l && tree[now].r <= r) {
tree[now].tag = (tree[now].tag * k) % P;
tree[now].lazy = (tree[now].lazy * k) % P;
tree[now].sum = (tree[now].sum * k) % P;
return ;
}
if(tree[now].tag != 1 || tree[now].lazy) pushdown(now);
ll mid = (tree[now].r + tree[now].l) >> 1;
if(mid >= l) muladd(l,r,k,now << 1);
if(mid < r) muladd(l,r,k,now << 1 | 1);
updata(now);
}
ll query(ll l,ll r,ll now) {
if( tree[now].l >= l && tree[now].r <= r )
return tree[now].sum % P;
ll mid = (tree[now].l + tree[now].r) >> 1;
ll Ans = 0;
if(tree[now].tag != 1 || tree[now].lazy) pushdown(now);
if(mid >= l) Ans += query(l,r,now << 1);
if(r > mid) Ans += query(l,r,now << 1 | 1);
return Ans % P;
}
void addadd(ll l,ll r,ll k,ll now) {
if(tree[now].l >= l && tree[now].r <= r) {
tree[now].lazy = (tree[now].lazy + k) % P;
tree[now].sum = (tree[now].sum + (tree[now].r - tree[now].l + 1) * k ) % P;
return;
}
if(tree[now].tag != 1 || tree[now].lazy) pushdown(now);
ll mid = (tree[now].l + tree[now].r) >> 1;
if(mid >= l) addadd(l,r,k,now << 1);
if(mid < r) addadd(l,r,k,now << 1 | 1);
updata(now);
return;
}
}stree;
int main() {
n = read();P = read();
for(ll i = 1;i <= n;++ i)
a[i] = read();
stree.build(1,1,n);
m = read();
for(ll i = 1,opt,l,r;i <= m;++ i) {
opt = read();l = read();r = read();
if(opt == 1) {
ll x = read();
stree.muladd(l,r,x,1);
}
if(opt == 2) {
ll x = read();
stree.addadd(l,r,x,1);
}
if(opt == 3)
printf("%d
",stree.query(l,r,1));
}
return 0;
}
2208: [Jsoi2010]连通数 (bitset)
(bitset)优化
#include <bitset>
#include <cstdio>
#include <iostream>
const int X = 2005;
std::bitset<X>g[X];
char z[X];
int main(){
int n;
scanf("%d",&n);
for(int i = 0;i < n;++ i){
scanf("%s",z);
for(int j = 0;j < n;++ j)
g[i][j] = (z[j] == '1' || i == j);
}
for(int k = 0;k < n;++ k){
for(int i = 0;i < n;++ i){
if(g[i][k])g[i] |= g[k];
}
}
int s = 0;
for(int i = 0;i < n;++ i)
s += g[i].count();
printf("%d
",s);
}
2875: [Noi2012]随机数生成器 (矩阵快速幂)
矩阵快速幂 + 快速乘
#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
ll m,a,c,x0,n,g;//模数 a c 第0项 n 第n项模除g
struct emm{
ll s[4][4];
emm() {memset(s,0,sizeof(s));}
}ans;
long long mul(long long a,long long b)
{
long long res=0;
for(;b;b>>=1)
{
if(b&1)res=(res+a)%m;
a=(a<<1)%m;
}
return res;
}
emm operator * (const emm &a,const emm &b) {
emm c;
for(ll i = 1;i <= 2;++ i)
for(ll j = 1;j <= 2;++ j)
for(ll k = 1;k <= 2;++ k)
c.s[i][j] = ( c.s[i][j] + mul(a.s[i][k] , b.s[k][j])) % m;
return c;
}
void fast_pow() {
emm now;
ans.s[1][1] = ans.s[2][2] = 1;
now.s[1][1] = a;
now.s[2][1] = c;
now.s[2][2] = 1;
for(;n;n >>= 1,now = now * now) {
if(n & 1) ans = ans * now;
}
emm c;
c.s[1][1] = x0;c.s[1][2] = 1;
c = c * ans;
std::cout << c.s[1][1] % g;
}
int main() {
std::cin >> m >> a >> c >> x0 >> n >> g;
fast_pow();
return 0;
}
3224: Tyvj 1728 普通平衡树 (splay)
splay模板
#include <bits/stdc++.h>
const int N = 1e6 + 10;
int ch[N][3], fa[N], size[N], cnt[N], key[N];
int Size, Root;
#define gc getchar()
inline int read() {
int x = 0, f = 1; char c = gc;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = gc;}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
return x * f;
}
void Clear(int x) {
ch[x][1] = ch[x][0] = cnt[x] = size[x] = fa[x] = key[x] = 0;
}
void Update(int x) {
if (x) {
size[x]=cnt[x];
if (ch[x][0]) size[x]+=size[ch[x][0]];
if (ch[x][1]) size[x]+=size[ch[x][1]];
}
}
bool Get(int x) {
return ch[fa[x]][1] == x;
}
void Rotate(int x) {
int fax = fa[x], grfa = fa[fax], w = Get(x);
ch[fax][w] = ch[x][w ^ 1];
fa[ch[x][w ^ 1]] = fax;
ch[x][w ^ 1] = fax;
fa[fax] = x;
fa[x] = grfa;
ch[grfa][ch[grfa][1] == fax] = x;
Update(fax);
Update(x);
}
void Splay(int x) {
for(int fax; fax = fa[x]; Rotate(x))
if(fa[fax]) Rotate((Get(x) == Get(fax)) ? fax : x);
Root = x;
}
void Insert(int x) {
if(! Root) {
Size ++;
Root = Size;
ch[Root][1] = ch[Root][0] = fa[Root] = 0;
size[Root] = cnt[Size] = 1;
key[Root] = x;
return ;
}
int now = Root, fanow = 0;
while(1) {
if(x == key[now]) {
cnt[now] ++;
Update(now), Update(fanow);
Splay(now);
return ;
}
fanow = now;
now = ch[now][key[now] < x];
if(!now) {
Size ++;
ch[Size][0] = ch[Size][1] = 0;
fa[Size] = fanow;
ch[fanow][x > key[fanow]] = Size;
size[Size] = cnt[Size] = 1;
key[Size] = x;
Update(fanow);
Splay(Size);
return;
}
}
}
int Pre() {
int now = ch[Root][0];
while(ch[now][1]) now = ch[now][1];
return now;
}
int Nxt() {
int now = ch[Root][1];
while(ch[now][0]) now = ch[now][0];
return now;
}
inline int Find(int x) {
int now = Root, Ans = 0;
while(1) {
if(x < key[now]) now = ch[now][0];
else {
Ans += (ch[now][0] ? size[ch[now][0]] : 0);
if(x == key[now]) {
Splay(now);
return Ans + 1;
}
Ans += cnt[now];
now = ch[now][1];
}
}
}
inline int Findx(int x) {
int now = Root;
while(1) {
if(ch[now][0] && x <= size[ch[now][0]]) now = ch[now][0];
else {
int imp = (ch[now][0] ? size[ch[now][0]] : 0) + cnt[now];
if(x <= imp) return key[now];
x -= imp;
now = ch[now][1];
}
}
}
void Delete(int x) {
int my = Find(x);
if(cnt[Root] > 1) cnt[Root] --, Update(Root);
else if(!ch[Root][1] && !ch[Root][0]) Clear(Root), Root = 0;
else if(!ch[Root][0]) {
int befroot = Root;
Root = ch[Root][1];
fa[Root] = 0;
Clear(befroot);
}
else if(!ch[Root][1]) {
int befroot = Root;
Root = ch[Root][0];
fa[Root] = 0;
Clear(befroot);
}
else {
int left = Pre(), befroot = Root;
Splay(left);
ch[Root][1] = ch[befroot][1];
fa[ch[befroot][1]] = Root;
Clear(befroot);
Update(Root);
}
return ;
}
int main() {
int T = read();
for(; T; T --) {
int opt = read(), x = read();
if(opt == 1) Insert(x);
else if(opt == 2) Delete(x);
else if(opt == 3) std:: cout << Find(x) << "
";
else if(opt == 4) std:: cout << Findx(x) << "
";
else if(opt == 5) {
Insert(x);
std:: cout << key[Pre()] << "
";
Delete(x);
} else {
Insert(x);
std:: cout << key[Nxt()] << "
";
Delete(x);
}
}
return 0;
}
3555: [Ctsc2014]企鹅QQ (hash)
hash
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define base 233
#define ull unsigned long long
using namespace std;
const int maxN = 30000 + 7;
ull hashe[maxN][200 + 7];
ull hs[maxN];
ull pw[maxN];
char s[maxN][200 + 7];
int n,L;
void init() {
pw[0] = 1;
for(int i = 1;i <= L;++ i)
pw[i] = pw[i - 1] * base;
}
int main() {
int tmp;
scanf("%d%d%d",&n,&L,&tmp);
init();
for(int i = 1;i <= n;++ i)
scanf("%s",s[i] + 1);
for(int i = 1;i <= n;++ i)
for(int j = 1;j <= L;++ j)
hashe[i][j] = hashe[i][j - 1] * base + s[i][j];
int k = 0;
for(int i = 1;i <= L;++ i) {
for(int j = 1;j <= n;++ j) {
hs[j] = hashe[j][i - 1] * pw[L - i] + hashe[j][L] - hashe[j][i] * pw[L - i];
}
bool flag = 0;
sort(hs + 1,hs + n + 1);
int t = 1;
for(int j = 2;j <= n;++ j) {
if(hs[j] != hs[j - 1]) t = 1;
else {
k += t ++;
}
}
}
printf("%d",k);
return 0;
}
4034: [HAOI2015]树上操作 (树链剖分)
树链剖分模板题
#include <iostream>
#include <algorithm>
#include <cstdio>
const int maxN = 1e5 + 7;
#define ll long long
using namespace std;
inline ll read() {
ll x = 0,f = 1;char c = getchar();
while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
return x * f;
}
ll n,m,R;
ll a[maxN];
struct Node{
ll l,r,w,lazy;
}tree[maxN << 2];
void pushdown(ll now) {
ll lazy = tree[now].lazy;
tree[now].lazy = 0;
tree[now << 1].lazy = (tree[now << 1].lazy + lazy );
tree[now << 1 | 1].lazy = (tree[now << 1 | 1].lazy + lazy) ;
tree[now << 1].w += (tree[now << 1].r - tree[now << 1].l + 1) * lazy ;
tree[now << 1 | 1].w += (tree[now << 1 | 1].r - tree[now << 1 | 1].l + 1) * lazy ;
return;
}
void updata(ll now) {
tree[now].w = (tree[now << 1].w + tree[now << 1 | 1].w) ;
}
void build(ll l,ll r,ll now) {
tree[now].l = l;tree[now].r = r;
if(l == r) return;
ll mid = (l + r ) >> 1;
build(l,mid,now << 1);
build(mid + 1,r,now << 1 | 1);
}
void Inter_add(ll l,ll r,ll now,ll val) {
if(tree[now].l >= l && tree[now].r <= r) {
tree[now].w += (tree[now].r - tree[now].l + 1) * val;
tree[now].lazy += val;
return;
}
pushdown(now);
ll mid = (tree[now].l + tree[now].r) >> 1;
if(l <= mid) Inter_add(l,r,now << 1,val);
if(r > mid) Inter_add(l,r,now << 1 | 1,val);
updata(now);
return;
}
ll query(ll l,ll r,ll now) {
if(tree[now].l >= l && tree[now].r <= r) {
return tree[now].w ;
}
if(tree[now].lazy)pushdown(now);
ll Ans = 0;
ll mid = (tree[now].l + tree[now].r) >> 1;
if(l <= mid) Ans += query(l,r,now << 1) ;
if(r > mid ) Ans += query(l,r,now << 1 | 1);
return Ans ;
}
void change(ll pos,ll val,ll now) {
if(tree[now].l == tree[now].r) {
tree[now].w = val ;
return;
}
ll mid = (tree[now].l + tree[now].r) >> 1;
if(pos <= mid) change(pos,val,now << 1);
else change(pos,val,now << 1 | 1);
updata(now);
return ;
}
struct qode {
ll v,nex;
}map[maxN << 1];
ll size[maxN],pos[maxN],top[maxN],dep[maxN],fa[maxN],maxsize,num;
ll head[maxN];
void add(ll u,ll v) {
map[ ++ num].v = v;
map[num].nex = head[u];
head[u] = num;
return;
}
void dfs1(ll u) {
size[u] = 1;
for(ll i = head[u];i;i = map[i].nex) {
ll v = map[i].v;
if(v == fa[u]) continue;
fa[v] = u;dep[v] = dep[u] + 1;
dfs1(v);
size[u] += size[v];
}
}//预处理树的大小,父亲节点和深度.
void dfs2(ll u,ll top1) {
ll k = 0;
maxsize ++;
pos[u] = maxsize;
top[u] = top1;
for(ll i = head[u];i;i = map[i].nex) {
ll v = map[i].v;
if(dep[v] > dep[u] && size[v] > size[k]) k = v;
}
if(!k) return;
dfs2(k,top1);
for(ll i = head[u];i;i = map[i].nex) {
ll v = map[i].v;
if(dep[v] > dep[u] && v != k) dfs2(v,v);
}
return;
}//求重儿子,在线段树中的位置.
void sloveadd(ll u,ll v,ll val) {
while(top[u] != top[v]) {
if(dep[top[u]] < dep[top[v]]) swap(u,v);
Inter_add(pos[top[u]],pos[u],1,val);
u = fa[top[u]];
}
if(pos[u] > pos[v]) swap(u,v);
Inter_add(pos[u],pos[v],1,val);
return ;
}//区间修改
ll slovequery(ll u,ll v) {
ll sum = 0;
while(top[u] != top[v]) {
if(dep[top[u]] < dep[top[v]]) swap(u,v);
sum += query(pos[top[u]],pos[u],1);
u = fa[top[u]];
}
if(pos[u] > pos[v]) swap(u,v);
sum += query(pos[u],pos[v],1);
return sum;
}//区间查询.
int main() {
n = read();m = read();
for(int i = 1;i <= n;++ i)
a[i] = read();
for(ll i = 1,u,v;i < n;++ i) {
u = read();v = read();
add(u,v);add(v,u);
}
R = 1;
dfs1(R);dfs2(R,R);
build(1,maxsize,1);
for(ll i = 1;i <= n;++ i)
change(pos[i],a[i],1);
for(ll i = 1,opt;i <= m;++ i) {
opt = read();
if(opt == 1) {
ll u,val;
u = read();val = read();
Inter_add(pos[u],pos[u],1,val);
}
if(opt == 3) {
ll u,v;
u = read();
printf("%lld
",slovequery(u,1));
}
if(opt == 2) {
ll x,z;
x = read();z = read();
Inter_add(pos[x],pos[x] + size[x] - 1,1,z);
}
}
return 0;
}