Uva12545

题目链接

如果S串中的’1’数量大于T串中的’1’数量那么肯定无法得到串T，所以在有解得情况下只要考虑S[i]=’1’&&T[i]=’0’和S[i]=’0’&&T[i]=’1’这两种情况的最大值再加上S串中为’？’的数量

#include <iostream>  
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <string>
#include <string.h>
#include <cmath>
#include <sstream>
#include <set>
#include <map>
#include <functional>
#include <queue>
#include <vector>
using namespace std;

const int maxn = 200;
int t;
char str1[maxn], str2[maxn];

int main()
{
    scanf("%d", &t);
    int Case = 0;
    getchar();
    while (t--)
    {
        int s1, t1, ss1, ss;
        s1 = t1 = ss1 = ss = 0;

        scanf("%s %s", str1, str2);
        int len = strlen(str1);
        for (int i = 0; i < len; i++)
        {
            if (str1[i] == '1' && str2[i] == '0')
                s1++;
            if (str1[i] == '0' && str2[i] == '1')
                t1++;
            if (str1[i] == '?')
                ss++;
            if (str1[i] == '?'&&str2[i] == '1')
                ss1++;          
        }
        if (s1 > t1 + ss1)
            printf("Case %d: -1\n", ++Case);
        else
            printf("Case %d: %d\n",++Case, max(s1, t1) + ss);
    }
}