TEST

test1

测试

test2

void printf_subset(int s){  //二进制状态下输出有1的位置下标
	for(int i = 0;i < n;i++){
		if(s & (1 << i))printf("%d ",i);
	}
	printf("\n");
}
void Comb(int k){ //枚举{0,...,n-1}中数量为k的子集
	int comb = (1 << k) - 1;
	while (comb < 1 << n) {	
        printf_subset(comb);
	    int x = comb & -comb, y = comb + x;
	    comb = ((comb&~y) / x >> 1) | y;
	}
}