思考 假设第一天买m朵,则m(2k - 1) = n,因为2 <= k <= 15,因此提前将2k - 1在2 <= k <= 15的值打印出来,然后for循环遍历,看n % (2k - 1) == 0 是否存在,如果存在输出YE5,否则输出N0代码如下: #include<iostream> #include<cstdio> int a[]={0,0,3,7,15,31,63,127,255,511,1023,2047,4095,8191,16383,32767}; int main() { int T; scanf("%d...