http://acm.hdu.edu.cn/showproblem.php?pid=2177

威佐夫博奕,面对奇异局势既bk=ak+k时是必败点,其中bk>=ak,k=bk-ak

别的处理和其他博弈相同,注意题目的数据取一堆的时候数据有问题

#include <iostream>
#include <cstdio>
#include <cmath>
#include <map>

using namespace std;
 
int OK(int b, int k) {
    if(b == (int)(k * (1 + sqrt(5.0)) / 2) + k) return 1;
    return 0;
}
 
int main() {
    int a, b;
    while(~scanf("%d%d", &a, &b)) {
        if(!a && !b) break;
        if(a > b) swap(a, b);
        int k = b - a;
        if(OK(b, k)) puts("0");
        else {
            puts("1");
            int x, y;
            for(int i = 1; i <= a; i++) {
                x = a - i, y = b - i;
                if(OK(y, y-x)) printf("%d %d\n", x, y);
            }
            map <int, int> mp;
            for(int i = 1; i <= b; i++) {
                x = a - i, y = b;
                if(x > 0 && OK(y, y-x) && mp[x]!=y) {
                    printf("%d %d\n", x, y);
                    mp[x] = y;
                }
                x = a, y = b - i;
                if(x > y) swap(x, y);
                if(OK(y, y-x) && mp[x]!=y) {
                    printf("%d %d\n", x, y);
                    mp[x] = y;
                }
            }
        }
    }
    return 0;
}
View Code