BZOJ 1874 取石子游戏

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 + 5;

int n, m, a[105], b[N];
int SG[N], S[105];

int read() {
    int X = 0, p = 1; char c = getchar();
    for(;c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}

void init() {
    for(int i = 0; i <= 1000; ++i) {
        int tmp = 0;
        memset(S, 0, sizeof(S));
        for(int j = 1; j <= m && b[j] <= i; ++j) 
            S[SG[i - b[j]]] = 1;
        for(; S[tmp]; ++tmp);
        SG[i] = tmp;
    }
}

int main()
{
    n = rd;
    for(int i = 1; i <= n; ++i)
        a[i] = rd;
    m = rd;
    for(int i = 1; i <= m; ++i)
        b[i] = rd;
    init();
    int ans = 0;
    for(int i = 1; i <= n; ++i)
        ans ^= SG[a[i]];
    if(ans) printf("YES
");
    else return printf("NO
"), 0;
    for(int i = 1; i <= n; ++i) {
        int tmp = ans ^ SG[a[i]];
        for(int j = 1; j <= m && b[j] <= a[i]; ++j)
            if((tmp ^ SG[a[i] - b[j]]) == 0)
                return printf("%d %d
", i, b[j]);
    }
}