字符串 || Gym 101653T Runes

给T个式子，其中有一个数字是？，让确定？代表哪个数字

负号，前导0，出现过的数字不能是？

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
char s[33];
int vis[11];
bool checkzero(int t)
{
    if(s[t] == ‘-‘) t++;
    if((s[t] == ‘?‘) && ((s[t+1] >= ‘0‘ && s[t+1] <= ‘9‘) || s[t+1] == ‘?‘) ) return 1; //no
    return 0;
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s", s);
        int len = strlen(s);
        long long a, b, c;
        memset(vis, 0, sizeof(vis));
        int op, idx, zf;
        for(int i = 0; i < len; i++)
        {
            if(s[i] >= ‘0‘ && s[i] <= ‘9‘) vis[s[i] - ‘0‘] = 1;//等式中出现过的数不能是?
            if(s[i] == ‘=‘) idx = i;
        }
        int ans, flag = 0;
        for(int j = 0; j <= 9; j++)//枚举?的所有可能情况
        {
            if(vis[j] == 1) continue;
            a = 0, b = 0, c = 0;
            zf = 1;//负号坑
            for(int i = 0; i < len; i++)
            {
                if(s[i] == ‘-‘ && i == 0) zf = -1;
                else if(s[i] == ‘+‘ || s[i] == ‘-‘ || s[i] == ‘*‘)
                {
                    op = i;
                    break;
                }
                else if(s[i] == ‘?‘) a = a * 10 + j;
                else a = a * 10 + s[i] - ‘0‘;
            }
            if(j == 0 && checkzero(0)) continue;
            a = a * zf;
            zf = 1;
            for(int i = op + 1; i < idx; i++)
            {
                if(s[i] == ‘-‘ && i == op+1) zf = -1;
                else if(s[i] == ‘?‘) b = b * 10 + j;
                else b = b * 10 + s[i] - ‘0‘;
            }
            if(j == 0 && checkzero(op+1)) continue;
            b = b * zf;
            zf = 1;
            for(int i = idx + 1; i < len; i++)
            {
                if(s[i] == ‘-‘ && i == idx+1) zf = -1;
                else if(s[i] == ‘?‘) c = c * 10 + j;
                else c = c * 10 + s[i] - ‘0‘;
            }
            if(j == 0 && checkzero(idx+1)) continue;
            c = c * zf;
            if(s[op] == ‘+‘ && a + b == c) {ans = j; flag = 1; break;}
            if(s[op] == ‘-‘ && a - b == c) {ans = j; flag = 1; break;}
            if(s[op] == ‘*‘ && a * b == c) {ans = j; flag = 1; break;}
        }
        if(flag) cout << ans << endl;
        else cout << "-1" << endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/pinkglightning/p/8978024.html