BZOJ 1874 取石子游戏 - SG函数

Description

$N$堆石子， $M$种取石子的方式， 最后取石子的人赢， 问先手是否必胜

$A_i <= 1000$,$ B_i <= 10$

Solution

由于数据很小， 直接暴力求SG函数即可判断。

Code

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define rd read()
 using namespace std;

 const int N = 1e3 + ;

 int n, m, a[], b[N];
 int SG[N], S[];

 int read() {
     int X = , p = ; char c = getchar();
     for(;c > '' || c < ''; c = getchar()) if(c == '-') p = -;
     for(; c >= '' && c <= ''; c = getchar()) X = X *  + c - '';
     return X * p;
 }

 void init() {
     for(int i = ; i <= ; ++i) {
         int tmp = ;
         memset(S, , sizeof(S));
         for(int j = ; j <= m && b[j] <= i; ++j)
             S[SG[i - b[j]]] = ;
         for(; S[tmp]; ++tmp);
         SG[i] = tmp;
     }
 }

 int main()
 {
     n = rd;
     for(int i = ; i <= n; ++i)
         a[i] = rd;
     m = rd;
     for(int i = ; i <= m; ++i)
         b[i] = rd;
     init();
     int ans = ;
     for(int i = ; i <= n; ++i)
         ans ^= SG[a[i]];
     if(ans) printf("YES\n");
     else return printf("NO\n"), ;
     for(int i = ; i <= n; ++i) {
         int tmp = ans ^ SG[a[i]];
         for(int j = ; j <= m && b[j] <= a[i]; ++j)
             if((tmp ^ SG[a[i] - b[j]]) == )
                 return printf("%d %d\n", i, b[j]);
     }
 }