【习题 8-3 UVA

【链接】 我是链接,点我呀:)
【题意】

在这里输入题意

【题解】

如果1的个数第一个串比第2个串多。 那么就无解。 否则。 找几个位置去凑1 优先找'?'然后才是0的位置 剩余的全都用swap操作就好

【代码】

/*
  	1.Shoud it use long long ?
  	2.Have you ever test several sample(at least therr) yourself?
  	3.Can you promise that the solution is right? At least,the main ideal
  	4.use the puts("") or putchar() or printf and such things?
  	5.init the used array or any value?
  	6.use error MAX_VALUE?
  	7.use scanf instead of cin/cout?
  	8.whatch out the detail input require
*/
/*
    一定在这里写完思路再敲代码!!!
*/
#include <bits/stdc++.h>
using namespace std;

int T,num[2][2],n;
string s1,s2;

void change(){
    for (int i = 0;i < n;i++)
        if (s1[i]=='?' && s2[i]=='1'){
            s1[i] = '1';
            return;
        }
    for (int i = 0;i < n;i++)
        if (s1[i]=='0' && s2[i]=='1'){
            s1[i]= '1';
            return;
        }
    for (int i = 0;i < n;i++)
        if (s1[i]=='?'){
            s1[i] = '1';
            return;
        }
    for (int i = 0;i < n;i++)
        if (s1[i]=='0') {
            s1[i] = '1';
            return;
        }
}

int main(){
	#ifdef LOCAL_DEFINE
	    freopen("rush_in.txt", "r", stdin);
	#endif
	ios::sync_with_stdio(0),cin.tie(0);
    cin >> T;
    int kase= 0;
    while (T--){
        memset(num,0,sizeof num);
        cin >> s1 >> s2;
        n = s1.size();
        for (int i = 0;i < n;i++)
            if (s1[i]>='0' && s1[i] <='1')
                num[0][s1[i]-'0']++;
        for (int i = 0;i < n;i++)
            if (s2[i]>='0' && s2[i] <='1')
                num[1][s2[i]-'0']++;

        if (num[0][1]>num[1][1]){
            cout <<"Case "<<++kase<<": "<< -1 << endl;
            continue;
        }
        int ans = 0;
        while (num[0][1] < num[1][1]){
            ans++;
            change();
            num[0][1]++;
        }
        for (int i = 0;i < n;i++)
            if (s1[i]=='?') s1[i] = '0',ans++;

        for (int i = 0;i < n;i++)
        if (s1[i]=='1' && s1[i]!=s2[i]){
            ans++;
        }
        cout <<"Case "<<++kase<<": "<< ans << endl;
    }
	return 0;
}