Hdu-6266-hdu 6266 Hakase and Nano (game theory)

Test instructions

There were two people taking stones from a heap of n stones, one of whom could take two times, and the second one could only be taken once. Finally, the man who finished the victory.

Ideas:

type	Hakase First	after Hakase
1	W	L
1 1	W	W
1 1 1 (3n)	L	W
1 1 1 1 (3n+1)	W	L
1 1 1 1 1 (3n+2)	W	W

type	Hakase First	after Hakase
X 1 1	W	L
X Y 1	W	W
X Y Z	W	W
X 1 1 1	W	L
X Y 1 1	W	W
X Y Z 1	W	W
X 1 1 1 1	W	W
X Y 1 1 1	W	W

1. If n = 3t, if each stone heap is 1,a the initiator must lose (1 1 1). There is only one stone heap with a quantity greater than 1, then B initiator a must lose (X 1 1).

2. If n = 3t+1, if only a bunch of persimmon is greater than 1 (X 1 1 1). Then B initiator a must lose.

Write the code according to the conclusion:

Code:

#include<iostream>using namespace std;int main() {    int t, n, d;    scanf("%d", &t);    while(t--) {        bool flag = true;        int cnt = 0, a;        scanf("%d %d", &n, &d);        for(int i = 0; i < n; i++) {            scanf("%d", &a);             if(a >= 2)                cnt++;        }        int x = n % 3;        if(x == 0) {            if(cnt==0 && d==1)                flag = false;            if(cnt==1 && d==2)                flag = false;        }else if(x == 1){            if(cnt<=1 && d==2)                flag = false;        }        if(flag) printf("Yes\n");        else printf("No\n");    }    return 0;}

Hdu-6266-hdu 6266 Hakase and Nano (game theory)