题目链接：　　　　https://codeforces.com/problemset/problem/1141/C

 

 

C. Polycarp Restores Permutation
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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−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.


Examples


input
3
-2 1
output
3 1 2 


input
5
1 1 1 1
output
1 2 3 4 5 


input
4
-1 2 2
output
-1

 

 

题意：     输入整数n  ，在输入n-1个数，代表着目标数组后一个减前一个数组的差的值，　　即：　ans[i+1] - ans[i]

　　　　若存在则输出数组ans的值　　　　　　若不存在则输出-1

 

 

思路： 　暴力水题。。。。。。。。。。。。。。

 

#include<iostream>
#include<cstdio>
#include<ctime>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<stack>
#include<map> 
#include<algorithm>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define Mem0(x) memset(x,0,sizeof(x))
#define Mem1(x) memset(x,-1,sizeof(x))
#define MemX(x) memset(x,0x3f,sizeof(x))
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f;
const double pi=acos(-1.0);

ll ans[2000010],q[2000010],cnt[2000010],n;
int main()
{
	memset(cnt,0,sizeof(cnt));
	scanf("%lld",&n);
	ll tmp=0,l=0;
	for (int i=1;i<n;i++){
		scanf("%lld",&q[i]);
		l+=q[i]; 
		tmp=tmp+l;
	}
	ll temp=(n+1)*n/2;
	ll pp=temp-tmp;
	bool flag=false;
	temp=pp/n;
	for (int i=1;i<=n;i++){
		ans[i]=temp;
		if (temp<=0||temp>n||cnt[ans[i]]>=1){
			flag=true;
		}
		else{
			cnt[ans[i]]++;
		}
		temp=temp+q[i];
	}
	if (flag)
		printf("-1");
	else{
		for (int i=1;i<=n;i++){
			printf("%lld ",ans[i]);
		}
	}
	printf("\n");
	return 0;
}