# 13692 - Aftermath's Ideology

>author: Utin
###### tags: `string`

---

## Brief
需建表並加入參數q以免TLE
String Operation目前已知共有4種變化
13572 String Operations 1
13573 String Operations 2
13692 Aftermath’s Ideology
13805 Professor Bear’s Challenge

## Solution 0

#include <stdio.h>
#include <string.h>

int q;
char arr[1000001]; // input
char table[100001][26]; // alphabet table
int op[100001][3]; // operation queue

int main() {
    scanf("%s", arr);
    scanf("%d", &q);

    int len = strlen(arr);

    for (int j = 0; j < 26; j++) table[q][j] = 'a' + j;

    // input
    for (int i = 0; i < q; i++) {
        scanf("%d", &op[i][0]);
        if (op[i][0] == 1) {
            char temp1, temp2;
            scanf(" %c %c", &temp1, &temp2);
            op[i][1] = temp1;
            op[i][2] = temp2;
        }
        else if (op[i][0] == 2) {
            char temp;
            scanf("%d %c", &op[i][1], &temp);
            op[i][2] = temp;
        }
        else scanf("%d %d", &op[i][1], &op[i][2]);
    }

    for (int i = q-1; i >= 0; i--) {
        // update
        for (int j = 0; j < 26; j++) {
            table[i][j] = table[i+1][j];
        }
        // operation type 1
        if (op[i][0] == 1) {
            table[i][op[i][1]-'a'] = table[i+1][op[i][2]-'a'];
        }
    }

    for (int i = 0; i < q; i++) {
        // operation type 2
        if (op[i][0] == 2) {
            arr[op[i][1]] = table[i][op[i][2]-'a'] - 32;
        }
        // operation type 3
        else if (op[i][0] == 3) {
            char temp = arr[op[i][1]];
            arr[op[i][1]] = arr[op[i][2]];
            arr[op[i][2]] = temp;
        }
    }

    // output
    for (int i = 0; i < len; i++) {
        if (arr[i] >= 'a') printf("%c", table[0][arr[i]-'a']);
        else printf("%c", arr[i] + 32);
    }
    printf("\n");
}
// By Utin

## Reference