CF 1110 E. Magic Stones

E. Magic Stones

链接

题意:

  给定两个数组,每次可以对一个数组选一个位置i($2 \leq i \leq n - 1$),让a[i]=a[i-1]+a[i+1]-a[i],或者b[i]=b[i-1]+b[i+1]-b[i]。问进行一些操作后,a和b能否相同。

分析:

  考虑一次操作会变成什么样子。

  a b c

  a a+c-b c

  我们发现这些数字差分后是不变的。于是对两个数组差分后,排序,看是否一样即可。注意判一下1和n是否相等。

代码:

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 100005;
LL a[N], b[N], c1[N], c2[N];

int main() {
    int n = read();
    for (int i = 1; i <= n; ++i) a[i] = read();
    for (int i = 1; i <= n; ++i) b[i] = read();
    if (a[1] != b[1] || a[n] != a[n]) {
        puts("No"); return 0;
    }
    for (int i = 1; i < n; ++i) c1[i] = a[i + 1] - a[i];
    for (int i = 1; i < n; ++i) c2[i] = b[i + 1] - b[i];
    sort(c1 + 1, c1 + n);
    sort(c2 + 1, c2 + n);
    for (int i = 1; i < n; ++i) {
        if (c1[i] != c2[i]) {
            puts("No"); return 0; 
        }
    }
    puts("Yes");
    return 0;
}

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转载自www.cnblogs.com/mjtcn/p/10357192.html