题目链接:http://nyoj.top/problem/1277

  • 内存限制:64MB 时间限制:1000ms 特判: No

题目描述

XiaoMing likes mathematics, and he is just learning how to convert numbers between different bases , but he keeps making errors since he is only 6 years old. Whenever XiaoMing converts a number to a new base and writes down the result, he always writes one of the digits wrong. For example , if he converts the number 14 into binary (i.e., base 2), the correct result should be "1110", but he might instead write down "0110" or "1111". XiaoMing never accidentally adds or deletes digits, so he might write down a number with a leading digit of " 0" if this is the digit she gets wrong. Given XiaoMing 's output when converting a number N into base 2 and base 3, please determine the correct original value of N (in base 10). (N<=10^10) You can assume N is at most 1 billion, and that there is a unique solution for N. 

输入描述

The first line of the input contains one integers T, which is the nember of test cases (1<=T<=8)
Each test case specifies:
* Line 1: The base-2 representation of N , with one digit written incorrectly.
* Line 2: The base-3 representation of N , with one digit written incorrectly. 

输出描述

For each test case generate a single line containing a single integer , the correct value of N

样例输入

1
1010
212

样例输出

14

解题思路

题意就是二进制串和三进制串都有一位是错误的，找出错误使其两个串相等。直接枚举两个串的每一位，直到满足条件使其相等。

#include <bits/stdc++.h>
using namespace std;
int main() {
    char a[1005], b[1005];
    int t, A, B, la, lb, temp;    
    scanf("%d%*c", &t);
    while (t--) {
        temp = 0;
        scanf("%s%s", a, b);
        la = strlen(a), lb = strlen(b);
        for (int i = 0; i < la && !temp; i++) {
            A = 0;
            a[i] = (a[i] - '0' + 1) % 2 + '0';
            for (int j = 0; j < la; j++)
                A = A * 2 + a[j] - '0';
            a[i] = (a[i] - '0' + 1) % 2 + '0';
            for (int j = 0; j < lb && !temp; j++) {
                for (int k = 0; k < 2 && !temp; k++) {
                    B = 0;
                    b[j] = (b[j] - '0' + 1) % 3 + '0';
                    for (int l = 0; l < lb; l++)
                        B = B * 3 + b[l] - '0';
                    if (A == B)
                        temp = 1;
                }
                b[j] = (b[j] - '0' + 1) % 3 + '0';
            }
        }
        printf("%d\n", A);
    }
    return 0;
}