【zzuli 1928 Decimal integer conversion】+ 模拟 + map-优快云博客

本文介绍了一个算法问题，即根据一位错误的二进制和三进制表示找出原始十进制数。通过分析输入的二进制和三进制字符串，算法能够定位并修正错误的数字，从而准确还原出原始数值。

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Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 18 Solved: 7

SubmitStatusWeb Board
Description

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.
Input

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.
Output

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

1
1010
212
Sample Output

14
HINT

Source

河南省第九届大学生程序设计竞赛

假设其中一串字符的某一个字符为错，对比另一个，注意：首位置不能为零

AC代码：

#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
typedef long long LL;
char s1[110],s2[110];
map <LL,LL> m;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
       // m.clear();
        scanf("%s %s",s1,s2);
        int n1 = strlen(s1),n2 = strlen(s2);
        if(s1[0] == '0'){
            s1[0] = '1';
            LL ans = 0,pl = 1;
            for(int i = n1 - 1; i >= 0; i--)
                ans += (s1[i] - '0') * pl,pl *= 2;
            printf("%lld\n",ans);
        }
        else if(s2[0] == '0'){
            s2[0] = '1';
            LL ans = 0,pl = 1;
            for(int i = n2 - 1; i >= 0; i--)
                ans += (s2[i] - '0') * pl,pl *= 3;
            m[ans]++;
            s2[0] = '2';
            ans = 0, pl = 1;
            for(int i = n2 - 1; i >= 0; i--)
                ans += (s2[i] - '0') * pl,pl *= 3;
            m[ans]++;
            ans = 0,pl = 1;
            for(int i = n1 - 1; i >= 0; i--)
                ans += (s1[i] - '0') * pl,pl *= 2;
            pl = 1;
           for(int i = n1 - 1; i >= 0; i--){
                if(s1[i] == '0' && m[ans + pl]) {
                    printf("%lld\n",ans + pl) ;break;
                }
                else if(i != 0 && m[ans - pl]) {
                        printf("%lld\n",ans - pl); break;
                }
                pl *= 2;
            }
        }
        else{
            LL ans = 0,pl = 1;
            for(int i = n2 - 1; i >= 0; i--) ans += (s2[i] - '0') * pl,pl *= 3;
            pl = 1;
            for(int i = n2 - 1; i >= 0; i--,pl *= 3){
                if(s2[i] == '0') m[ans + pl]++,m[ans + pl * 2]++;
                else if(s2[i] == '1'){
                      if(i != 0)  m[ans - pl]++;
                      m[ans + pl]++;
                }
                else {
                    m[ans - pl]++;
                    if(i != 0) m[ans - 2 * pl]++;
                }
            }
            ans = 0,pl = 1;
            for(int i = n1 - 1; i >= 0; i--)
                ans += (s1[i] - '0') * pl,pl *= 2;
            pl = 1;
            for(int i = n1 - 1; i >= 0; i--){
                if(s1[i] == '0' && m[ans + pl]) {
                    printf("%lld\n",ans + pl) ;break;
                }
                else if(i != 0 && m[ans - pl]) {
                        printf("%lld\n",ans - pl); break;
                }
                pl *= 2;
            }
        }
    }
    return 0;
}