F - Polycarp Restores Permutation CodeForces - 1141C

最新推荐文章于 2020-09-05 11:39:15 发布

原创最新推荐文章于 2020-09-05 11:39:15 发布 · 201 阅读

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本文介绍了一种算法，用于帮助用户根据给定的差值数组恢复一个遗忘的整数排列。通过使用前缀和与映射(map)技巧，确保了所求排列的正确性和唯一性。文章分享了作者解决问题的心路历程，包括多次尝试与失败，最终在学长的帮助下找到了正确的解决方案。

An array of integers p1,p2,…,pnp1,p2,…,pn is called a permutation if it contains each number from 11 to nn exactly once. For example, the following arrays are permutations: [3,1,2][3,1,2], [1][1], [1,2,3,4,5][1,2,3,4,5] and [4,3,1,2][4,3,1,2]. The following arrays are not permutations: [2][2], [1,1][1,1], [2,3,4][2,3,4].

Polycarp invented a really cool permutation p1,p2,…,pnp1,p2,…,pn of length nn. It is very disappointing, but he forgot this permutation. He only remembers the array q1,q2,…,qn−1q1,q2,…,qn−1 of length n−1n−1, where qi=pi+1−piqi=pi+1−pi.

Given nn and q=q1,q2,…,qn−1q=q1,q2,…,qn−1, help Polycarp restore the invented permutation.

Input

The first line contains the integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the length of the permutation to restore. The second line contains n−1n−1 integers q1,q2,…,qn−1q1,q2,…,qn−1 (−n<qi<n−n<qi<n).

Output

Print the integer -1 if there is no such permutation of length nn which corresponds to the given array qq. Otherwise, if it exists, print p1,p2,…,pnp1,p2,…,pn. Print any such permutation if there are many of them.

3
-2 1

3 1 2

5
1 1 1 1

1 2 3 4 5

4
-1 2 2

-1

之前想求出q1，后来一直wa加re。。。（从37到42），后来换了一种思路，用前缀求和（也re，到47），就在网上看到可以用map。果然ac了。。。最后学长帮我看了一下，交换了一下一个地方就ac了。。。以后细心一点，判断的地方应该先进行哪些判断

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<cstring>
#include<string>
#include<math.h>
using namespace std;
typedef long long ll;
int main()
{
	ll n,q,flag=1,x[200005];
	ll minn=0;
	bool y[200005];
	x[0]=0;
	memset(y,false,sizeof(y));
	scanf("%lld",&n);
	for(int i=1;i<n;i++)
	{
		scanf("%lld",&q);
		x[i]=x[i-1]+q;
		minn=min(minn,x[i]);
	}
	for(int i=0;i<n;i++)
	{
		if(x[i]-minn+1>n||y[x[i]-minn+1]==true)
		{
			flag=0;
			break;
		}
		y[x[i]-minn+1]=true;
	}
	if(flag==0) printf("-1\n");
	else
	{
		for(int i=0;i<n-1;i++) printf("%lld ",x[i]-minn+1);
		printf("%lld\n",x[n-1]-minn+1);
	}
	return 0;
}