分析

这个可以暴力求出一个循环内的变换。这样可以考虑置换乘法的结合率，然后把置换拆成循环。这样是可以通过的。但是其实现在等同于只有一个操作重复 次，这个其实可以倍增优化，时间复杂度为 。

代码

#include<bits/stdc++.h>
using namespace std;
int read() {
    int x = 0,f = 0;char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-')f = 1;ch = getchar();}
    while(isdigit(ch)) {x = x * 10 + ch - '0';ch = getchar();}
    return f ? -x : x;
}
#define LL long long 
const LL N = 1e6+100;
LL n,m,k,B[N],A[N],f[N][40];
int main()
{
    n = read();m = read();k = read();
    for(int i = 1;i <= n;i++) A[i] = i;
    while(m--) {
        int l = read(),r = read();
        memcpy(B,A,sizeof(B));
        for(int i = 1;i <= r-l+1;i++) {
            B[r-i+1] = A[l+i-1];
        }
        memcpy(A,B,sizeof(A));
    }
    for(int i = 1;i <= n;i++) f[i][0] = A[i];
    for(int j = 1;j <= 35;j++) {
        for(int i = 1;i <= n;i++) {
            f[i][j] = f[f[i][j-1]][j-1];
        }
    }
    for(int i = 1;i <= n;i++) {
        int x = i;
        for(int j = 35;~j;j--) {
            if((k>>j)&1) x = f[x][j];
        }
        printf("%d ",x);
    }
    printf("\n");
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int read() {
    int x = 0,f = 0;char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-')f = 1;ch = getchar();}
    while(isdigit(ch)) {x = x * 10 + ch - '0';ch = getchar();}
    return f ? -x : x;
}
const int inf = 0x3f3f3f3f,N = 1e6+100;
int n,m,k,B[N],A[N],pos[N],vis[N],len[N],st[N],top = 0;
int lcm = 1;
int main()
{
    n = read();m = read();k = read();
    for(int i = 1;i <= n;i++) A[i] = i;
    while(m--) {
        int l = read(),r = read();
        memcpy(B,A,sizeof(B));
        for(int i = 1;i <= r-l+1;i++) {
            B[r-i+1] = A[l+i-1];
        }
        memcpy(A,B,sizeof(A));
    }
    for(int i = 1;i <= n;i++){
        int loop = 0;
        int t = i;
        while(!vis[t]){
            loop++;
            st[++top] = t;
            vis[t] = true;
            t = B[t];

        }
        while(top) len[st[top--]] = loop; 
    }
    for(int i = 1;i <= n;i++) {
        int c = k%len[i],pos = i;
        while(c--) pos = B[pos];
        printf("%d ",pos);
    }
    return 0;
}