A. Game with chocolates
因为差值必须是$P$的幂，故首先可以$O(log n)$枚举出先手第一步所有取法，判断之后的游戏是否先手必败。
对于判断，首先特判非法的情况，并假设$n<m$，则题意可理解成将$n$或者$m$减小至$n-P^k$，在$P$进制下可以理解为$n$某一位减$1$，且这一位在减之前不能是$0$。
若是将$m$减小为$n-P^k$，则整个游戏都是确定的，回合数为$n$的数位之和，根据奇偶性即可判断胜负。
但若是将$n$减小为$m-P^k$，则要求$n$和$m$位数相同且最高位相等，这意味着进行这步操作后之后不能再进行这一步操作，先手可以利用这一步来使自己必胜。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll P,n,m,k,a[99],b[99];
bool check(ll n,ll m){
if(n>m)swap(n,m);
ll k=m-n;
if(m/k%P==0)return 1;
while(k%P==0)k/=P;
if(k>1)return 1;
if(!n)return 0;
int la=0,lb=0;
while(n)a[++la]=n%P,n/=P;
while(m)b[++lb]=m%P,m/=P;
if(la==lb&&a[la]==b[lb])return 1;
for(int i=2;i<=la;i++)a[1]+=a[i];
return a[1]%2;
}
int main(){
scanf("%lld%lld%lld",&P,&n,&m);
for(k=1;k<=n;k*=P)if(n-k<m)if(!check(n,n-k)){
puts("YES");
printf("%lld %lld",n,n-k);
return 0;
}
for(k=1;k<=m;k*=P)if(m-k<n)if(!check(m-k,m)){
puts("YES");
printf("%lld %lld",m-k,m);
return 0;
}
puts("NO");
}
B. Birches
将相同的数合并，然后调和级数$O(nlog n)$枚举即可。
#include<cstdio>
const int N=111111;
int n,m,k,i,j,x,l,r,f[N];long long ans;
int main(){
scanf("%d%d",&n,&k);
while(n--){
scanf("%d",&x);
f[x]++;
}
n=100000;
for(i=1;i<=n;i++)if(f[i]&&k<i)
for(l=k;l<=n;l+=i)ans+=1LL*f[i]*f[l];
printf("%lld",ans);
}
C. Ancient CBS
按平方数分解构造。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
char s[(int)3e5];
int h, t;
void solve(int w, int st)
{
for(int i = st; w >= i; ++i)
{
s[t++] = '(';
s[t++] = ')';
w -= i;
}
if(w == 0)return;
s[--h] = '(';
s[t++] = ')';
if(--w == 0)return;
solve(w, 2);
}
int main()
{
while(~scanf("%d", &n))
//for(int x = 1, n = 1e9 - x; x <= 1000000; ++x, --n)
{
h = t = 1e5; s[t] = 0;
solve(n, 1); s[t] = 0;
puts(s + h);
/*printf("%d
", t - h);
if(t - h > 1e5)
{
puts("NO");
while(1);
}*/
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
*/
D. Interactive lock
爆搜得出方案即可。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
int a[100] = {0,
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 22, 24, 14, 18, 16, 17, 19, 20, 21, 23, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 78, 44, 45, 46, 47, 48, 49, 50, 51, 100, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 98, 92, 97, 99, 96};
int top = 96;
/*
bool e[N];
int p[N];
int v[N];
void prime()
{
int g = 0;
int k = 0;
for(int i = 2; i <= 10000; ++i)
{
if(!e[i])
{
p[++g] = i;
for(int j = i + i; j <= 10000; j += i)
{
e[j] = 1;
}
}
else
{
v[++k] = i;
}
}
}
*/
set<int>can;
bool dfs(int g, vector<int>rst)
{
if(rst.size() == 0)
{
puts("bingo");
printf("%d
", g - 1);
for(int i = 1; i < g; ++i)printf("%d, ", a[i]); puts("");
return 1;
}
set<int>::iterator it, nit;
for(it = can.begin(); it != can.end(); it = nit)
{
if(*it > rst.front())return 0;
vector<int>nxt;
for(auto x : rst)if(x % *it)
{
nxt.push_back(x - *it);
}
a[g] = *it;
nit = it; ++nit;
can.erase(a[g]);
if(dfs(g + 1, nxt))return 1;
can.insert(a[g]);
}
/*for(auto it : can)
{
if(it > rst.front())return 0;
vector<int>nxt;
for(auto x : rst)if(x % it)
{
nxt.push_back(x - it);
}
a[g] = it;
can.erase(a[g]);
if(dfs(g + 1, nxt))return 1;
can.insert(a[g]);
}*/
return 0;
}
void solve()
{
for(int i = 2; i <= 100; ++i)can.insert(i);
vector<int>rst;
for(int i = 100; i <= 10000; ++i)rst.push_back(i);
dfs(1, rst);
}
bool guess(int x)
{
for(int i = 1; i <= top; ++i)
{
if(a[i] > x)
{
printf("%d
", x);
return 0;
}
if(x % a[i] == 0)return 1;
x -= a[i];
}
return 0;
}
int main()
{
//prime();
//solve();
int T; scanf("%d", &T);
while(T--)
{
for(int i = 1; i <= top; ++i)
{
printf("%d
", a[i]); fflush(stdout);
char s[10]; scanf("%s", s);
if(s[0] == 'Y')break;
}
}
/*
for(int i = 100; i <= 10000; ++i)
{
if(!guess(i))
{
printf("%d
", i);
}
}
puts("haha");
*/
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
*/
E. Interval divisibility
对于每个约数计算贡献，分段求和即可。
时间复杂度$O(sqrt{n})$。
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
const ll P=1000000007,inv2=(P+1)/2;
ll f(ll n){
n%=P;
return n*(n+1)%P*inv2%P;
}
ll cal(ll n){
ll ret=0;
for(ll i=1;i<=n;){
ll j=n/(n/i);
ret+=f(n/i)*((i+j)%P)%P*((j-i+1)%P)%P*inv2%P;
ret%=P;
i=j+1;
}
return ret;
}
int main(){
ll l,r;
cin>>l>>r;
ll ans=cal(r)-cal(l-1);
ans=ans%P;
ans=ans+P;
ans%=P;
cout<<ans;
}
F. A trick
分类讨论。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
int main()
{
while(~scanf("%d", &n))
{
if(n == 0)
{
puts("-1");
continue;
}
int x = n;
int sum = 0;
while(x)
{
sum += x % 10;
x /= 10;
}
if(sum == 9 * 9)
{
puts("-1");
}
else
{
int tmp1 = sum;
int v1 = 0;
while(tmp1)
{
int can = min(tmp1, 9);
tmp1 -= can;
v1 = v1 * 10 + can;
}
int tmp2 = sum;
int v2 = 0;
bool flag = 1;
while(tmp2)
{
int can = min(tmp2, 9 - flag);
flag = 0;
tmp2 -= can;
v2 = v2 * 10 + can;
}
if(v1 != n)printf("%d
", v1);
else if(v2 != n)printf("%d
", v2);
else printf("%d0
", v1);
}
}
return 0;
}
/*
【trick&&吐槽】
【题意】
【分析】
【时间复杂度&&优化】
*/
G. Highest ratings year
首先求出所有路径长度之和，并直接除以$2$，那么奇数长度的路径会算错，故再算出奇数长度的路径数即可。
时间复杂度$O(n)$。
#include<cstdio>
typedef long long ll;
const int N=100010;
int n,i,x,y,g[N],v[N<<1],nxt[N<<1],ed;
int cnt[2],d[N],size[N];
ll ans;
inline void add(int x,int y){v[++ed]=y;nxt[ed]=g[x];g[x]=ed;}
void dfs(int x,int y){
d[x]=d[y]^1;
cnt[d[x]]++;
size[x]=1;
for(int i=g[x];i;i=nxt[i])if(v[i]!=y){
dfs(v[i],x);
size[x]+=size[v[i]];
ans+=1LL*size[v[i]]*(n-size[v[i]]);
}
}
int main(){
scanf("%d",&n);
for(i=1;i<n;i++)scanf("%d%d",&x,&y),add(x,y),add(y,x);
dfs(1,0);
ll odd=1LL*cnt[0]*cnt[1];
ans/=2;
ans+=odd-odd/2;
printf("%lld",ans);
}
H. Spells
设$f[i][j]$表示考虑$S$前$i$个位置，被一路交换过来的字符是$j$能匹配成功的字符串个数，当且仅当$j$和当前字符不相同时才进行转移。
那么转移可以写成矩阵的形式，快速幂计算。
对于矩阵的构造，注意到每次单个字符的转移矩阵与$E$相比只有常数个位置不同，故可以利用这点在$O(26)$时间内计算矩阵乘法。
时间复杂度$O(26sum|S|+n26^3log k)$。
#include<cstdio>
#include<cstring>
#define rep(i) for(int i=0;i<N;i++)
const int N=28,P=1000000007;
int n,m,K,f[N],g[N][N];char s[10010];
inline void mul(){
static int h[N][N];
rep(i)rep(j)h[i][j]=0;
rep(i)rep(j)if(g[i][j])rep(k)if(g[j][k])h[i][k]=(1LL*g[i][j]*g[j][k]+h[i][k])%P;
rep(i)rep(j)g[i][j]=h[i][j];
}
inline void apply(){
static int h[N];
rep(i){
h[i]=0;
rep(j)h[i]=(1LL*g[i][j]*f[j]+h[i])%P;
}
rep(i)f[i]=h[i];
}
inline void gao(int x){//2..27
//w[x][0]=1
//w[x][x]=P-1
//w[0][1]=1
//w[0][x]=P-1
//w[1][x]=P-1
//w[1][0]=1
static int h[N][N];
rep(i){
h[0][i]=g[0][i];
h[1][i]=g[1][i];
h[x][i]=g[x][i];
}
rep(i){
g[x][i]=(h[0][i]+g[x][i])%P;
g[x][i]=(1LL*(P-1)*h[x][i]+g[x][i])%P;
g[0][i]=(h[1][i]+g[0][i])%P;
g[0][i]=(1LL*(P-1)*h[x][i]+g[0][i])%P;
g[1][i]=(1LL*(P-1)*h[x][i]+g[1][i])%P;
g[1][i]=(h[0][i]+g[1][i])%P;
}
}
int main(){
scanf("%d",&n);
f[0]=1;
//1:sum = 2+...+27
while(n--){
scanf("%s%d",s,&K);
m=strlen(s);
rep(i)rep(j)g[i][j]=i==j;
for(int i=0;i<m;i++)gao(s[i]-'a'+2);
for(;K;K>>=1,mul())if(K&1)apply();
}
printf("%d",f[0]);
}
/*
7
a 1
n 1
g 1
n 1
g 1
a 1
n 1
6
a 1
n 1
g 1
n 1
g 1
an 1
*/
I. Silver table
$n$的方案为将$n-1$的方案复制后放在上下左右四个地方，并将右上块和左下块全体加$2^k-1$，再将左下块反对角线全体加$2^k-1$得到。
#include<cstdio>
const int N=3222;
int i,j,k,n,f[N][N];
int main(){
f[1][1]=1;
for(i=2;i<=11;i++){
int len=1<<(i-1);
int hal=len/2;
for(j=1;j<=hal;j++)for(k=1;k<=hal;k++){
f[j+hal][k+hal]=f[j][k];
f[j+hal][k]=f[j][k+hal]=f[j][k]+len-1;
}
for(j=1;j<=hal;j++)f[j+hal][j]+=len-1;
}
scanf("%d",&n);
n=1<<n;
for(i=1;i<=n;i++){
for(j=1;j<=n;j++)printf("%d ",f[i][j]);
puts("");
}
}
J. Soldier’s life
问题等价于找两条间距最小的平行线夹住所有点，故求出凸包后枚举每条边求出最远点即可。
#include<cstdio>
#include<algorithm>
#include<set>
#include<cmath>
using namespace std;
typedef double DB;
const int N=10000;
const DB eps = 1e-9, pi = acos(-1.0);
DB ans=1e100;
int n,m,i;
struct PT{
DB x, y;
PT(DB x = 0, DB y = 0):x(x), y(y){}
void input(){scanf("%lf%lf", &x, &y);}
PT operator-(const PT&p)const{return PT(x-p.x,y-p.y);}
PT operator+(const PT&p)const{return PT(x+p.x,y+p.y);}
PT operator*(double p)const{return PT(x*p,y*p);}
PT operator/(double p)const{return PT(x/p,y/p);}
bool operator < (const PT &p) const{
if(fabs(x - p.x) > eps) return x < p.x; return y < p.y;}
void output(){printf("%.15f %.15f
", x, y);}
DB len()const{return hypot(x,y);}
PT rot90()const{return PT(-y,x);}
PT trunc(double l)const{return (*this)*l/len();}
}a[N],b[N],c[N],fina,finb,f[N],fin[N];
DB lim=1e100;
DB cross(const PT&a,const PT&b){return a.x*b.y-a.y*b.x;}
DB vect(PT p, PT p1, PT p2){
return (p1.x - p.x) * (p2.y - p.y) - (p1.y - p.y) * (p2.x - p.x);
}
int convex_hull(PT *p, int n, PT *q){
int i, k, m; sort(p, p + n); m = 0;
for(i = 0; i < n; q[m++] = p[i++])
while(m > 1 && vect(q[m - 2], q[m - 1], p[i]) < eps) m --;
k = m;
for(i = n - 2; i >= 0; q[m ++] = p[i --])
while(m > k && vect(q[m - 2], q[m - 1], p[i]) < eps) m --;
return --m;
}
void solve(PT A,PT B){
DB mx=0;
for(int i=0;i<n;i++)mx=max(mx,fabs(cross(a[i]-A,B-A)));
mx/=(B-A).len();
mx/=2;
if(mx<ans){
ans=mx;
fina=A;
finb=B;
}
}
PT line_intersection(PT a,PT b,PT p,PT q){
double U=cross(p-a,q-p),D=cross(b-a,q-p);
return a+(b-a)*(U/D);
}
void gao(PT A,PT B){
PT fa=A-B;
fa=fa.rot90();
DB ret=0;
for(int i=0;i<n;i++){
PT C=a[i],D=C+fa;
PT now=line_intersection(A,B,C,D);
DB dis=(now-C).len();
ret=max(ret,dis);
f[i]=now;
}
if(ret<lim){
lim=ret;
for(int i=0;i<n;i++)fin[i]=f[i];
}
}
int main(){
scanf("%d",&n);
for(i=0;i<n;i++)a[i].input(),b[i]=a[i];
m=convex_hull(b,n,c);
for(i=0;i<m;i++)solve(c[i],c[(i+1)%m]);
PT tmp=finb-fina;
tmp=tmp.rot90();
tmp=tmp.trunc(ans);
gao(fina+tmp,finb+tmp);
gao(fina-tmp,finb-tmp);
printf("%.10f
",lim);
for(i=0;i<n;i++)fin[i].output();
}
K. Casino
DP，$f[n][m]$表示还剩$n$张红卡，$m$张黑卡的最大期望收益。
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=111;
int n,m;double f[N][N];bool v[N][N];
double dfs(int n,int m){
if(!n&&!m)return 0;
if(v[n][m])return f[n][m];
v[n][m]=1;
double ret=0;
if(n)ret+=(dfs(n-1,m)+1)*n;
if(m)ret+=(dfs(n,m-1)-1)*m;
return f[n][m]=max(ret/(n+m),0.0);
}
int main(){
scanf("%d%d",&n,&m);
printf("%.15f",dfs(n,m));
}