AcWing 2058. 笨拙的手指（暴力枚举）

Posted 2022-02-11 MangataTS

题目链接

https://www.acwing.com/problem/content/2060/

思路

因为二进制和三进制有一位是错误的，那么我们直接二重循环枚举每一位，看更改后是否满足相等，如果是的话那么这就是答案，否则继续枚举

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl "\\n"

ll ans;

bool fg(string a,string b) 
    int aa = 0;
    int bb = 0;
    for(int i = a.size()-1,j = 0; i >= 0; --i,j++) 
        aa += (a[i]-'0') * (1<<j);
    
    for(int i = b.size()-1,j = 1;i >= 0; --i,j*=3)
        bb += (b[i]-'0') * (j);
    
    ans = aa;
    return aa==bb;


bool check(int la,int lb,string a, string b)
    a[la]=a[la]=='0'?'1':'0';
    int k = b[lb]-'0';
    for(int i = 0;i <= 2; ++i) 
        if(i == k) continue;
        b[lb] = i + '0';
        if(fg(a,b)) return true;
    
    return false;


int main()

    string a,b;
    cin>>a>>b;
    int n = a.size();
    int m = b.size();
    for(int i = 0;i < n; ++i) 
        for(int j = 0;j < m; ++j) 
            if(check(i,j,a,b))
                cout<<ans<<endl;
                return 0;
            
        
    
    
    
    return 0;