判断n种操作能否使n个人胜负概率都为50%
简单思维题,即n-1次里一半胜一般负,所以n-1为偶数,即n为奇数才成立
#include<cstdio>
#include<cmath>
int main(){
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
if(n&1)
printf("Balanced\n");
else
printf("Bad\n");
}
return 0;
}