标签：要求 algo long cli exist targe div XML ant
Grigory has nn magic stones, conveniently numbered from 11 to nn . The charge of the ii -th stone is equal to cici .
Sometimes Grigory gets bored and selects some inner stone (that is, some stone with index ii , where 2≤i≤n?12≤i≤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 cici changes to c′i=ci+1+ci?1?cici′=ci+1+ci?1?ci .
Andrew, Grigory‘s friend, also has nn stones with charges titi . 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 cici into titi for all ii ?
The first line contains one integer nn (2≤n≤1052≤n≤105 ) — the number of magic stones.
The second line contains integers c1,c2,…,cnc1,c2,…,cn (0≤ci≤2?1090≤ci≤2?109 ) — the charges of Grigory‘s stones.
The second line contains integers t1,t2,…,tnt1,t2,…,tn (0≤ti≤2?1090≤ti≤2?109 ) — the charges of Andrew‘s stones.
If there exists a (possibly empty) sequence of synchronization operations, which changes all charges to the required ones, print "Yes".
Otherwise, print "No".
4 7 2 4 12 7 15 10 12
Yes
3 4 4 4 1 2 3
No
In the first example, we can perform the following synchronizations (11 -indexed):
In the second example, any operation with the second stone will not change its charge.
思路：
注意看题目给出的公式
c[i]‘=c[i+1]+c[i-1]-c[i]
可以变形成
c[i+1]-c[i]‘=c[i]-c[i-1]
c[i]‘-c[i-1]=c[i+1]-c[i]
这样看来就是将第i项左右差分交换，可以把这个看成一种排序
所以将差分全部重新排序之后，如果上下差分都相同，则满足题目要求
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
ll a[maxn],b[maxn],c[maxn],d[maxn];
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++) scanf("%I64d",&a[i]);
for(int i=1;i<=n;i++) scanf("%I64d",&b[i]);
for(int i=1;i<n;i++)
{
c[i]=a[i+1]-a[i];
d[i]=b[i+1]-b[i];
}
sort(c+1,c+n);
sort(d+1,d+n);
if(a[1]!=b[1]||a[n]!=b[n])
{
printf("No\n");
return 0;
}
for(int i=1;i<n;i++)
{
if(c[i]!=d[i])
{
printf("No\n");
return 0;
}
}
printf("Yes\n");
return 0;
}
标签：要求 algo long cli exist targe div XML ant
原文地址：https://www.cnblogs.com/EchoZQN/p/10357305.html
