CF L. Lexicographically Minimal Poem

最新推荐文章于 2024-09-23 03:49:01 发布

原创最新推荐文章于 2024-09-23 03:49:01 发布 · 1.4k 阅读

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CC 4.0 BY-SA版权

本文详细解析并提供了代码实现，涉及字符串操作、字符映射、数据结构应用等核心算法优化，旨在提高编程效率与理解深度。

尼玛，写的博客又被csdn吞了，早晚有一天我会换博客的，fuck！

不写了，直接贴代码。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
#include <set>
#include <cmath>
using namespace std;
typedef long long LL;
typedef pair<string,string> P;
const int maxn = 1000000 + 5;
const int INF = 1000000000;

char c[30];
char re[30];
char s[maxn];
int len[maxn];
vector<char> V[maxn];
vector<char> ans;

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        cin >> s >> c;
        for(int i = 0;i < 26;i++){
            re[c[i]-'a'] = i + 'a';
        }
        for(int i = 0;i < maxn;i++) V[i].clear();
        ans.clear();
        for(int i = 0;i < n;i++){
            char tem = s[i];
            for(int j = 1;j <= 26;j++){
                tem = re[tem-'a'];
                V[i].push_back(tem);
                if(tem == s[i]){
                    len[i] = j;
                    break;
                }
            }
        }
        int a = 0,b = 1;
        for(int i = 0;i < n;i++){
            char Min = 'z';
            int k = 0;
            int vis = a%len[i];
            int missyou;
            for(int j = 0;j < len[i]+1;j++){
                int pos = a + b*j;
                if(j > 0 && pos%len[i] == vis){
                    missyou = j;
                    break;
                }
                char tem = V[i][pos%len[i]];
                if(tem < Min){
                    Min = tem;
                    k = j;
                }
            }
            a += b*k;
            b *= missyou;
            ans.push_back(Min);
        }
        for(int i = 0;i < n;i++) printf("%c",ans[i]);
        printf("\n");
    }
    return 0;
}