AcWing 2058. 笨拙的手指（暴力枚举）

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

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

#include
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<= 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<