CF1110E Magic Stones - 差分

Posted 2020-12-01 lrj124

题面

Grigory has n n magic stones, conveniently numbered from (1) to (n). The charge of the (i)-th stone is equal to (c_i).

Sometimes Grigory gets bored and selects some inner stone (that is, some stone with index (i), where (2 le i le n-1)), and after that synchronizes it with neighboring stones. After that, the chosen stone loses its own charge, but acquires the charges from neighboring stones. In other words, its charge (c_i) changes to (c_i' = c_{i + 1} + c_{i - 1} - c_i)

Andrew, Grigory‘s friend, also has n n stones with charges (t_i). Grigory is curious, whether there exists a sequence of zero or more synchronization operations, which transforms charges of Grigory‘s stones into charges of corresponding Andrew‘s stones, that is, changes (c_i) into (t_i) for all (i)?

题意

给定两个数组 (c) 和 (t)，一次操作可以把 (c_i) 变成 (c_{i-1}+c_{i+1}-c_{i} (2 le i le n-1))，问若干次操作后，可不可以把 (c) 数组变成 (t)

思路

设 (a) 为 (c) 的差分数组（即 (a_i=c_i-c_{i-1})）
每对 (c_i) 进行一次操作后：
[a_i=c_i-c_{i-1}=(c_{i+1}-c_{i-1}-c_i)-c_{i-1}=c_{i+1}-c_i=a_{i+1}]
[a_{i+1}=c_{i+1}-c{i}=c_{i+1}-(c_{i+1}-c_{i-1}-c_i)=c_i-c_{i-1}=a_{i}]
所以，每次操作会将 (c) 的差分数组交换两个数，只要判一下 (c) 与 (t) 的差分数组是否相同即可

代码

#include <bits/stdc++.h>
using namespace std;
int n,a[100001],b[100001],c[100001],d[100001];
int main() {
    cin >> n;
    for (int i = 1;i <= n;i++) {
        cin >> a[i];
        if (i ^ 1) b[i-1] = a[i]-a[i-1];
    }
    sort(b+1,b+n);
    for (int i = 1;i <= n;i++) {
        scanf("%d",&c[i]);
        if (i ^ 1) d[i-1] = c[i]-c[i-1];
    }
    sort(d+1,d+n);
    if (a[1] ^ c[1] || a[n] ^ c[n]) {
        printf("No");
        return 0;
    }
    for (int i = 1;i < n;i++)
        if (b[i] ^ d[i]) {
            printf("No");
            return 0;
        }
    printf("Yes");
    return 0;
}