CF1110E Magic Stones

题目链接：（http://codeforces.com/problemset/problem/1110/E?csrf_token=c50ced6ad87d78ffc77d786512a94249 ）

E. Magic Stones
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Grigory has n magic stones, conveniently numbered from 1 to n. The charge of the i-th stone is equal to ci.
Sometimes Grigory gets bored and selects some inner stone (that is, some stone with index i, where 2≤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 ci changes to c′i=ci+1+ci−1−ci.
Andrew, Grigory’s friend, also has n stones with charges ti. 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 ci into ti for all i?
Input
The first line contains one integer n (2≤n≤105) — the number of magic stones.
The second line contains integers c1,c2,…,cn (0≤ci≤2⋅109) — the charges of Grigory’s stones.
The second line contains integers t1,t2,…,tn (0≤ti≤2⋅109) — the charges of Andrew’s stones.
Output
If there exists a (possibly empty) sequence of synchronization operations, which changes all charges to the required ones, print “Yes”.
Otherwise, print “No”.

解法：

说一下自己在解这一道题的时候出现的问题：
想法上的错误：

对于差分序列的转换，求出两个差分序列中的数字di，di+1，应该是始终只变换ci这一个数字，这样子的话就可以发现在ci变换过之后就是di和di+1变了一下位置，但是差分序列始终没有变化，所以，根据这一个发现，就可以通过二者差分序列在排序之后是否相等判断ci是否可以通过变化变成ti。
对于开头来说，没有ci-1，结尾也不能操作，应该也是要判的。但是自己认为结尾是直接包含在差分序列中的，这个思想是错误的，结尾从意义看是不能操作的，只要不能操作，那维护就没有意义，也是不能维护的。

code

#include<bits/stdc++.h>
using namespace std;const int nn=1e5+10;int n;
int c[nn],t[nn];
int b1[nn],b2[nn];inline int read()
{int x=0,f=1; char ch=getchar();while(ch>'9'||ch<'0'){if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;
}int main()
{n=read();for(int i=1;i<=n;++i) c[i]=read(),b1[i]=c[i]-c[i-1];for(int i=1;i<=n;++i)    t[i]=read(),b2[i]=t[i]-t[i-1];if(c[1]!=t[1]||c[n]!=t[n]){printf("No\n");return 0;}sort(b1+1,b1+1+n);sort(b2+1,b2+1+n);for(int i=1;i<=n;++i)if(b1[i]!=b2[i]){printf("No\n");return 0;}printf("Yes\n");return 0;
}

今天也有进步哦，加油！

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CF1110E Magic Stones

CF1110E Magic Stones

解法：

code

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