AGC006C Rabbit Exercise

Posted 2021-02-08 cjoieryl

传送门
设 (f_{i,j}) 表示兔子 (i) 在当前 (j) 轮的期望位置
对于一次操作 (f_{i,j+1}=frac{1}{2}(2f_{i-1,j}-f_{i,j})+frac{1}{2}(2f_{i+1,j}-f_{i,j})=f_{i-1,j}+f_{i+1,j}-f_{i,j})
这个东西就是差分数组上两个位置的交换
相当于是求经过 (k) 次长度为 (m) 的置换后的位置
只要求每个位置在每个环走 (k) 的位置即可

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
     
const int maxn(1e5 + 5);
     
int n, m, b[maxn], nxt[maxn], que[maxn], len, vis[maxn];
ll k, f[maxn], g[maxn];
     
int main() {
    int i, j;
    scanf("%d", &n);
    for (i = 1; i <= n; ++i) scanf("%lld", &f[i]), b[i] = i;
    for (i = n; i; --i) f[i] -= f[i - 1];
    scanf("%d%lld", &m, &k);
    for (i = 1; i <= m; ++i) scanf("%d", &j), swap(b[j], b[j + 1]);
    for (i = 1; i <= n; ++i)
        if (!vis[i]) {
            vis[i] = 1, j = b[i], que[len = 1] = i;
            while (!vis[j]) que[++len] = j, vis[j] = 1, j = b[j];
            for (j = 1; j <= len; ++j) nxt[que[j]] = que[(k + j - 1) % len + 1];
        }
    for (i = 1; i <= n; ++i) g[i] = f[nxt[i]];
    for (i = 1; i <= n; ++i) g[i] += g[i - 1], printf("%lld
", g[i]);
    return 0;
}