玲珑Round #7-1071 - Boring Game

1071 - Boring Game

Time Limit：1s Memory Limit：1024MByte

Submissions：612Solved：200

DESCRIPTION

In a mysterious cave, PigVan ( Mr.Van's pet ) has lived for thousands years. PigVan absorbed the power of nature, and it may pretend a human in speaking, walking and so on.

One day, he bought some valuable stone, and divided them into n piles of stone where the ith pile (1≤i≤n) contains valuesai .

After PigVan put them in a line, he wants to play a game.

In the boring game, he can do this operation:

Choose a stone pile ai （i>1）and its two adjacent piles ai-1, ai+1, turn(ai-1, ai, ai+1) to(ai-1 + ai, -ai, ai + ai+1).

PigVan wonders whether he can get (b1, b2, b3, …, bn) after several operations.

Note:

If you choose the last pile an, the operation will be( an-1 + an, -an ) .

INPUT

The first line is a single integer $T$, indicating the number of test cases.
For each test case:
In the first line, there are only one integer $n$(n≤105), indicating the number of food piles.
The second line is $n$ integers indicate sequence $a$ ( | ai | ≤ 106).
The third line is $n$ integers indicate sequence $b$ ( | bi | ≤ 106).

OUTPUT

For each test case, just print 'Yes' if PigVan can get$b$ after some operations; otherwise, print 'No'.

SAMPLE INPUT

261 6 9 4 2 0 7 -6 19 2 -6 641 2 3 4 4 2 1 3

SAMPLE OUTPUT

YesNo

SOLUTION

“玲珑杯”ACM比赛 Round #7

题解：(ai-1, ai, ai+1) -> (ai-1 + ai, -ai, ai + ai+1).

他们前后的和不变，只是顺序不同，所以只需要判断前缀和是否相等

code：

#include<cstdio>#include<algorithm>using namespace std;int main(){int a[100000],b[100000];int t,n;scanf("%d",&t);while(t--){scanf("%d",&n);a[0]=0;b[0]=0;for(int i=1;i<=n;i++){scanf("%d",&a[i]);a[i]+=a[i-1];}for(int i=1;i<=n;i++){scanf("%d",&b[i]);b[i]+=b[i-1];}sort(a+1,a+n+1);sort(b+1,b+n+1);int faut=1;for(int i=1;i<=n;i++){if(a[i]!=b[i])faut=0;}if(faut==1) printf("Yes\n");else printf("No\n");}return 0;}