seve bit take cas -s font print tween get

Diciption Let‘s consider m apples divided into n groups. Each group contains no more than 100 apples, arranged in a line. You can take any number of consecutive apples at one time.
For example "@@@" can be turned into "@@" or "@" or "@ @"(two piles). two people get apples one after another and the one who takes the last is
the loser. Fra wants to know in which situations he can win by playing strategies (that is, no matter what action the rival takes, fra will win).

Input

You will be given several cases. Each test case begins with a single number n (1 <= n <= 100), followed by a line with n numbers, the number of apples in each pile. There is a blank line between cases.
Output

If a winning strategies can be found, print a single line with "Yes", otherwise print "No".
Sample Input

2
2 2
1
3

Sample Output

No
Yes


anti_nim遊戲。
這種Nim是不能操作者贏。
Sg函數的話還是和普通的Nim一樣，初始化是sg[0]=0；
最後遊戲的和的話，先手必勝當且僅當：
    1.每一堆石子都是1並且Nim和為0.
    2.有至少一堆石子>1並且Nim和不為0.

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int sg[105],n,m;
int now,v[1005];

inline void init(){
	sg[0]=0;
	for(int i=1;i<=100;i++){
		now=0;
		for(int j=0;j<i;j++)
		    for(int k=j;k+j<i;k++) v[sg[j]^sg[k]]=i;
		while(v[now]==i) now++;
		
		sg[i]=now;
	}
}

int main(){
	init();
	
	while(scanf("%d",&n)==1){
		bool flag=0;
		int a,ans=0;
		
		while(n--){
			scanf("%d",&a);
			ans^=sg[a];
			if(a>1) flag=1;
		}
		
		if((!flag&&!ans)||(flag&&ans)) puts("Yes");
		else puts("No");
	}
	
	return 0;
}

　　

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