Polycarp Restores Permutation

https://codeforces.com/contest/1141/problem/C

题解：

a为给定数组

b为待求数组

b[i+1]=b[i]+a[i];

其中b[1]一定为零，当前的其他元素为相对于b[1]的相对差值相同与需要求的数组的与b[1]的相对差值；

因为b是一个排列，所以最小为1，找到b中最小的元素(并且小于等于0)，算出它与1的差值，其他元素都加上这个差值；

然后判断有木有重复和超过n的数。

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
int ans,cnt,flag,temp,sum;
int a[N];
int b[N];
bool vis[N];
char str;
struct node{};
void dfs(int x){
    if(x>=n){
        for(int i=1;i<=n;i++){
            printf("%d%c",b[i]," \n"[i==n]);
        }
        exit(0);
        return ;
    }
    b[x+1]=a[x]+b[x];//cout<<x<<" "<<b[x+1]<<endl;
    if(b[x+1]>0&&b[x+1]<=n&&vis[b[x+1]]==0){
        vis[b[x+1]]=1;
        dfs(x+1);
        vis[b[x+1]]=0;
    }
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);
    int minl=INF;
    for(int i=1;i<n;i++){
        scanf("%d",&a[i]);
        b[i+1]=a[i]+b[i];
        minl=min(minl,b[i+1]);
    }
    minl=min(minl,0);
    ans=abs(minl)+1;
    set<int>st;
    for(int i=1;i<=n;i++){
        b[i]+=ans;
        if(b[i]<=0||b[i]>n){
            cout<<-1<<endl;
            return 0;
        }
        st.insert(b[i]);
    }
    if(st.size()==n){
        for(int i=1;i<=n;i++){
            printf("%d%c",b[i]," \n"[i==n]);
        }
    }else{
        cout<<-1<<endl;
    }


    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}