Codeforces 484C Strange Sorting(置换)

题目链接：Codeforces 484C Strange Sorting

题目大意：给定一个长度为N的字符串，现在有M次询问，每次要从左向右逐个对长度为K的子串进行D-sorting，最后

输出生成的串。

解题思路：问题即为一个置换的思想，L对应的左移一位的置换，C对应的是D-sorting前K为的置换，每次执行完一次C

肯定执行一下L，保证D-sorting的为不同的K长度子串。用类似矩阵快速幂的思想对字符串进行求解，最后在有循环移

动对应的N-K位。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1e6+5;

int N, M, K, D, L[maxn];
int C[maxn], P[maxn], x[maxn], tmp[maxn];
char str[maxn];

void multi () {
    for (int i = 0; i < N; i++) tmp[i] = x[x[i]];
    for (int i = 0; i < N; i++) x[i] = tmp[i];
}

int main () {
    scanf("%s%d", str, &M);
    N = strlen(str);
    for (int i = 0; i < N; i++)
        L[i] = (i ? i - 1 : N - 1);

    while (M--) {
        scanf("%d%d", &K, &D);

        for (int i = 0; i < N; i++) C[i] = i;
        int mv = 0;
        for (int i = 0; i < D; i++) {
            for (int j = i; j < K; j += D)
                C[j] = mv++;
        }
        for (int i = 0; i < N; i++) P[i] = C[i];
        for (int i = 0; i < N; i++) x[i] = C[L[i]];

        int n = N - K;
        while (n) {
            if (n&1) {
                for (int i = 0; i < N; i++)
                    P[i] = x[P[i]];
            }
            multi();
            n >>= 1;
        }

        for (int i = 0; i < N; i++) tmp[P[i]] = str[i];
        for (int i = 0; i < N; i++) str[(i + (N - K)) % N] = tmp[i];
        printf("%s\n", str);
    }
    return 0;
}