Description
Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.
---Wikipedia
Today, Nim takes revenge on you. The rule of the game has changed a little: the player must remove the objects from the current head(first) heap. Only the current head heap is empty can the player start to remove from the new head heap. As usual, the player who takes the last object wins.
---Wikipedia
Today, Nim takes revenge on you. The rule of the game has changed a little: the player must remove the objects from the current head(first) heap. Only the current head heap is empty can the player start to remove from the new head heap. As usual, the player who takes the last object wins.
Input
The first line contains a single integer T, indicating the number of test cases.
Each test case begins with an integer N, indicating the number of heaps. Then N integer Ai follows, indicating the number of each heap successively, and the player must take objects in this order, from the first to the last.
[Technical Specification]
1. 1 <= T <= 100
2. 1 <= N <= 1 000
3. 1 <= Ai <= 1 000 000 000
Each test case begins with an integer N, indicating the number of heaps. Then N integer Ai follows, indicating the number of each heap successively, and the player must take objects in this order, from the first to the last.
[Technical Specification]
1. 1 <= T <= 100
2. 1 <= N <= 1 000
3. 1 <= Ai <= 1 000 000 000
Output
For each test case, output “Yes” if the first player can always win, otherwise “No”.
Sample Input
2 1 2 2 1 1
Sample Output
Yes No
思路： 博弈， 先找必胜状态或必败状态， 判断控制权在谁手里，他就必胜， 判断控制权在谁手里就判断第一个非1的数钱有几个1 如果是偶数个控制权就在第一个手里。
代码：
#include<stdio.h> #include<string.h> #define N 110000 int main(void) { int t, i, j, n, f, ok; int a[N]; scanf("%d", &t); while(t--) { f = 0; ok = 1; scanf("%d", &n); for(i = 0; i < n; i++) scanf("%d", &a[i]); for(i = 0; i < n; i++) { if(a[i] > 1) { ok = 0;//判断是否全是1. break; } else f++;//判断第一个不是1的位置。 } if(ok==0)//如果有非1的存在，就判断控制权再谁手里， 如果在第一非1的数前有偶数个1控制权在第一个人手里。 { if(f%2) printf("No "); else printf("Yes "); } else//如果全是1的话，如果有奇数堆第一个人必胜。 { if(f%2) printf("Yes "); else printf("No "); } } return 0; }