题面
T1
思路
考虑一下每个数会与其他位置的哪些数字遇到。显然每隔gcd(n,m,k)个数都会遇到一次。所以只要看一下将给出的所有数字全部对gcd(n,m,k)取模是否能包含从0到gcd(n,m,k) - 1的所有数就行了。
代码
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 100000 + 100;
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;
}
int n,m,K,a[N],b[N],k[N];
int gcd(int x,int y) {
return !y ? x : gcd(y,x % y);
}
namespace BF1 {
void Main() {
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
int aa = read();
for(int i = 1;i <= aa;++i) a[read()] = 1;
int bb = read();
for(int i = 1;i <= bb;++i) b[read()] = 1;
read();
for(int i = 0;i <= 100000;++i) {
int x1 = i % n, x2 = i % m;
a[x1] = b[x2] = a[x1] | b[x2];
}
for(int i = 0;i < n;++i) {
if(a[i] != 1) {
puts("No");
return;
}
}
for(int i = 0;i < m;++i) {
if(b[i] != 1) {
puts("No");
return;
}
}
puts("Yes");
}
}
namespace BF2 {
int tmp[N * 5],js = 0;
void Main() {
js = 0;
int mod = gcd(gcd(n,m),K);
int aa = read();
for(int i = 1;i <= aa;++i) tmp[++js] = read() % mod;
int bb = read();
for(int i = 1;i <= bb;++i) tmp[++js] = read() % mod;
int kk = read();
for(int i = 1;i <= kk;++i) tmp[++js] = read() % mod;
int now = 0;
sort(tmp + 1,tmp + js + 1);
tmp[0] = -1;
int ans = 0;
for(int i = 1;i <= js;++i) {
if(tmp[i] == tmp[i-1]) continue;
if(tmp[i] == tmp[i-1]+1) now++;
else {
ans = max(ans,now);
now = 1;
}
}
ans = max(ans,now);
if(ans >= mod) puts("Yes");
else puts("No");
return;
}
}
int main() {
freopen("happy2.in","r",stdin);
freopen("happy2.out","w",stdout);
int T = read();
while(T--) {
n = read(),m = read(),K = read();
if(n <= 100 && m <= 100 && K == 0) {
BF1::Main();
continue;
}
BF2::Main();
}
return 0;
}
T2
想了一会感觉不可做，直接55分暴力。
55分代码
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<cstring>
#include<algorithm>
#include<bitset>
using namespace std;
typedef long long ll;
const int N = 300000 + 100;
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;
}
int n,p;
namespace BF1 {
int du[N];
void Main() {
for(int i = 1;i <= n;++i) {
int x = read(), y = read();
du[x]++;
du[y]++;
}
ll ans1 = 0;
for(int i = 1;i <= n;++i)
if(du[i] == 0) ans1++;
cout<<(ll)n * (ll)(n - 1)/2 - (ans1 * (ans1 - 1) / 2);
return;
}
}
namespace BF2 {
const int NN = 110;
bitset <NN> tmp[NN];
int ans = 0;
inline int check(int x,int y) {
bitset <NN> ls;
ls = tmp[x] | tmp[y];
return ls.count() >= p;
}
void Main() {
for(int i = 1;i <= n;++i) {
int x = read(),y = read();
tmp[x][i] = 1;
tmp[y][i] = 1;
}
for(int i = 1;i <= n;++i) {
for(int j = i + 1;j <= n;++j) {
ans += check(i,j);
}
}
cout<<ans<<endl;
}
}
int main() {
freopen("suspect.in","r",stdin);
freopen("suspect.out","w",stdout);
n = read(),p = read();
if(p == 0) {
cout<<((ll)n*(n-1) / 2);
return 0;
}
if(p == 1) {
BF1::Main();
return 0;
}
if(n <= 100) {
BF2::Main();
return 0;
}
return 0;
}
std
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cstdlib>
#define INF 100000000
using namespace std;
typedef long long LL;
struct Edge
{
int from,to,pre;
}e[1000000];
int h[300005]={0},cou=0;
int c[300005],ed[300005];
void Addedge(int from,int to)
{
cou++;
e[cou]=((Edge){from,to,h[from]});
h[from]=cou;
}
inline void update(int x)
{
if(x==0)
{
c[0]++;
return;
}
for(;x<=300000;x+=x&-x)
c[x]++;
}
inline int get(int x)
{
if(x==-1) return 0;
int sum=0;
for(;x;x-=x&-x)
sum+=c[x];
return sum+c[0];
}
int main()
{
freopen("suspect.in","r",stdin);
freopen("suspect.out","w",stdout);
LL ans=0;
int sum,i,n,p,a,b,v,j;
cin>>n>>p;
for(i=1;i<=n;i++)
{
scanf("%d%d",&a,&b);
Addedge(a,b); Addedge(b,a);
ed[a]++; ed[b]++;
}
for(i=1;i<=n;i++)
update(ed[i]);
for(i=1;i<=n;i++)
{
if(ed[i]>=p)
ans+=n-1;
else
{
sum=n-get(p-ed[i]-1);
if(ed[i]>=p-ed[i]) sum--;
for(j=h[i];j;j=e[j].pre)
{
v=e[j].to;
if(ed[v]==p-ed[i]) sum--;
ed[v]--;
}
for(j=h[i];j;j=e[j].pre)
{
v=e[j].to;
ed[v]++;
}
ans+=sum;
}
}
cout<<ans/2<<endl;
return 0;
}
T3
80分思路
暴力分感觉都可做，然后就写了80分暴力。用莫队卡一下100分？？？会TLE吧(真麻烦，不想写)。
100分思路
如果这个题让着求区间出现奇数次的数的异或和就很简单了。所以我们可以对于询问先离线一下。按照右端点拍个序。然后用树状数组维护一下区间异或和。就是从1扫到n，同时将当前的数上一次出现的位置(在树状数组里)异或上这个数。然后查询就好了。
100分代码
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cstdlib>
#include<string>
#include<bitset>
#include<iomanip>
#include<deque>
#include<utility>
#define INF 1000000000
#define fi first
#define se second
#define N 1000005
#define P 1000000007
#define debug(x) cerr<<#x<<"="<<x<<endl
#define MP(x,y) make_pair(x,y)
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
int c[N],now,sum,a[N],b[N],ans[N],nxt[N],n;
map<int,int> vis,pre;
vector<pii> Q[N];
void Add(int x,int w)
{
for(;x<=n;x+=x&-x)
c[x]^=w;
}
int Get(int x)
{
int s=0;
for(;x;x-=x&-x)
s^=c[x];
return s;
}
int main()
{
int i,m,ql,qr,j;
freopen("xor.in","r",stdin);
freopen("xor.out","w",stdout);
cin>>n;
now=0;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
vis[a[i]]++;
nxt[pre[a[i]]]=i;
pre[a[i]]=i;
if(vis[a[i]]>1)
now^=a[i];
b[i]=now;
//debug(b[i]);
}
cin>>m;
for(i=1;i<=m;i++)
{
scanf("%d%d",&ql,&qr);
Q[ql].push_back(MP(qr,i));
}
for(i=1;i<=n;i++)
{
//debug(sum);
for(j=0;j<Q[i].size();j++)
ans[Q[i][j].se]=Get(Q[i][j].fi)^b[Q[i][j].fi];
ql=nxt[i];
if(ql)
{
sum^=a[i];
Add(ql,a[i]);
}
}
for(i=1;i<=m;i++)
printf("%d
",ans[i]);
return 0;
}
总结
期望得分:100 + 55 + 80 = 235
实际得分:100 + 55 + 80 = 235
暴力没挂真的感动。越来越菜了
一言
那时我怎么都想不到，原来也有这一天，念及你，竟既无风雨也无晴。 ——我亦飘零久