Codeforces Round #547 (Div. 3)C. Polycarp Restores Permutation_vasya has written some permutation p1p2…pnp1p2…pno-优快云博客

本文介绍了一种算法，用于帮助用户根据给定的差分数组恢复一个遗忘的整数排列。通过设定初始值并调整偏移量，确保最终排列符合1到n的范围。文章详细解释了算法的实现步骤，并提供了完整的C++代码示例。

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time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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−1integers 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.

Examples

input

Copy

3
-2 1

output

Copy

3 1 2

input

Copy

5
1 1 1 1

output

Copy

1 2 3 4 5

input

Copy

4
-1 2 2

output

Copy

-1

意义就是给了数组q，q是根据式子 qi=pi+1−pi，得到的然后让求原数组p。

这里先假设p1 = 1,也可以设成别的数，然后根据pi+1 = pi + qi，得到后续的数，然后又因为原数组的范围是1~n故，最小值应该是1,如果最小值不为1说明第一个元素也不是1，每个元素就有一个偏移量，（最小位置上的值 + 偏移量 = 1），偏移量 = 1 -

最小位置上的值，然后每个元素都加上偏移量即可得到原数组p;

#include<bits/stdc++.h>

using namespace std;
const int N = 2e5 + 10;
int p[N],a[N];

int main()
{
	int n;
	map<int,int> mmp; 
	scanf("%d",&n);
	for (int i=1;i<n;i++)
		scanf("%d",&p[i]);
	a[1] = 1;
	int temp = 0x3f3f3f3f; 
	for (int i=1;i<n;i++)
	{
		a[i+1] = p[i] + a[i];
		temp = min(temp,a[i+1]);
	}
	temp = min(temp,1);
	int k = 1 -  temp;
	for (int i=1;i<=n;i++)
	{
		a[i] += k;
		mmp[a[i]]++;
	}
	for (int i=1;i<=n;i++)
	{
		if (mmp[i]!=1)
		{
			printf("-1\n");
			return 0;
		}
	}
	for (int i=1;i<=n;i++)
		printf("%d ",a[i]);
	puts("");
	return 0;
}