题解 | 小红的字符串修改

#include <algorithm>
#include <cmath>
#include <iostream>
using namespace std;

//并不存在巧劲，最少的替换次数来源于暴力枚举。在不断比较下才得到最小值
int main() {
    string s_handle ;
    string s_target ;
    cin >> s_handle;
    cin >> s_target;
    int round = s_target.size() - s_handle.size();
    int minn = -1;
    for (int i = 0; i <= round; i++) { // 滑动窗口执行次数，计算每次的成本
        int summ = 0;
        for (int j = 0; j < s_handle.size(); j++) {
            summ += min(abs(s_target[j+i] - s_handle[j]), 26 - abs(s_target[j+i] - s_handle[j])); // z b 2, 26 - 2 = 24; a b 1,26 - 1 = 25
        }
        if (minn == -1|| minn > summ) {// 如果是-1，说明从未比较过，minn直接得到summ值
            minn = summ;
        }
    }
    cout << minn;
}
// 64 位输出请用 printf("%lld")