hdu 4662

将U全部转化为I   因为 I 的个数一定是2的n次方  有可能消除了一定数量的 2U  所以I的个数加上一个6的整数倍是2的n次方

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

char a[1000010];
int b[40] = {1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824};
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",a);
        int len = strlen(a);
        int sum = 0;
        if(a[0] == 'M')
        {
            int flag = 0,flag2 = 0;
            for(int i = 1; i < len; i++)
            {
                if(a[i] == 'I')
                {
                    sum++;
                }
                else if(a[i] == 'U')
                {
                    sum += 3;
                }
                else
                {
                    flag2 = 1;
                    break;
                }
            }
            if(!flag2)
            {
                for(int i = 0; i <= 29; i++)
                {
                    if(b[i] >= sum && (b[i]-sum) % 6 == 0)
                    {
                        flag = 1;
                        break;
                    }
                }
                if(flag)
                    puts("Yes");
                else
                    puts("No");
            }
            else
                puts("No");
        }
        else
            puts("No");
    }
    return 0;
}