2020 CCPC Wannafly Winter Camp Day1 Div.1&2 B.密码学

题目链接 https://ac.nowcoder.com/acm/contest/3979/B

题意：给你n个字符串，再给你m个操作，xi，yi， 意思是用是s[xi]来加密s[yi]，加密方法是(s[xi]+s[yi]) %52.加密后的替换原理的s[yi];

题解：先用一个二维数组存下所有的操作，然后再倒还原就可以了

#include<iostream>
#include<string>
#include <cstdlib>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include <iomanip>

// #pragma comment(linker, "/STACK:1024000000,1024000000")
// #define pi acos(-1)
// #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x7f7f7f7f //2139062143
#define INF1 0x3f3f3f3f //1061109567
#define INF2 2147483647
#define llINF 9223372036854775807
#define pi 3.141592653589793//23846264338327950254
#define pb push_back
#define ll long long
#define debug cout << "debug\n";
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
// ios::sync_with_stdio(false);cin.tie(NULL);
#define scai(x) scanf("%d", &x)
#define sca2i(x, y) scanf("%d %d", &x, &y)
#define scaf(x) scanf("%lf", &x)
#define sca2f(x, y) scanf("%lf %lf", &x, &y)
#define For(m,n) for (int i = m;  i < n; i++)

#define local
#ifdef local
#endif
const int N = 1e5 + 2;

int a[1000];
// 用数组的方式构建一个映射
void init()
{
    int j = 0;
    int i = 0;
    for (i = 'a'; i <= 'z'; i++, j++)
    {
        a[i] = j;
    }
    j = 'a';
    for (i = 0; i < 26; i++,j++)
    {
        a[i] = j;
    }
    j = 26;
    for (i = 'A'; i <= 'Z'; i++, j++)
    {
        a[i] = j;
    }
    j = 'A';
    for (i = 26; i < 52; i++,j++)
    {
        a[i] = j;
    }
}
string s[1005];
int op[1003][2];
int main()
{
    init();
    int i;
    /*for(i = 0; i < 123; i++)
        printf("%d %d\n", i, a[i]);*/
    int n, m;
    cin >> n>> m;

    for (int i = 0; i < m; i++)  // 先存下所有加密操作
    {
        cin >> op[i][0] >> op[i][1];
        op[i][0]--;
        op[i][1]--;
    }
    for (int i = 0; i < n; i++)
    {
        cin >> s[i];

    }
    for (int i = m - 1; i >= 0; i--)
    {
        string t1 = s[op[i][0]];
        string t2 = s[op[i][1]];
        //cout << t1 <<endl << t2 << endl;
        while (t1.size() < t2.size())     // 如果加密代码太短，就在后面复制一段
            t1.append(t1);
        int len2 = t2.size();
        for (int i = 0; i < len2; i++)
        {
            t2[i] = a[(a[t2[i]] + 104 - a[t1[i]]) %52];
           /*if (a[t2[i]] > a[t1[i]])
            {
                t2[i] = a[a[t2[i]] - a[t1[i]]];
            }
            else t2[i] = a[a[t2[i]] + 52 - a[t1[i]]];*/
        }
        s[op[i][1]] = t2.substr(0, t2.size()); // 留下与原长一样的字符串替换加密字符串
    }
    for (int i = 0; i < n; i++)
    {
        cout << s[i] << endl;
    }

}