題目：
Farmer John is lining up his N cows (2≤N≤103), numbered 1…N, for a photoshoot. FJ initially planned for the i-th cow from the left to be the cow numbered ai, and wrote down the permutation a1,a2,…,aN on a sheet of paper. Unfortunately, that paper was recently stolen by Farmer Nhoj!
Luckily, it might still possible for FJ to recover the permutation that he originally wrote down. Before the sheet was stolen, Bessie recorded the sequence b1,b2,…,bN−1 that satisfies bi=ai+ai+1 for each 1≤i<N.
Based on Bessie’s information, help FJ restore the “lexicographically minimum” permutation a that could have produced b. A permutation x is lexicographically smaller than a permutation y if for some j, xi=yi for all i<j and xj<yj (in other words, the two permutations are identical up to a certain point, at which x is smaller than y). It is guaranteed that at least one such a exists.
Input
The first line of input contains a single integer N.
The second line contains N−1 space-separated integers b1,b2,…,bN−1.
Output
A single line with N space-separated integers a1,a2,…,aN.
Example
inputCopy
5
4 6 7 6
outputCopy
3 1 5 2 4
Note
a produces b because 3+1=4, 1+5=6, 5+2=7, and 2+4=6.
這個題目非常之狗，看起來無從下手，但是隻要開始假設，因爲他是字典序最小的答案，所以你讓答案序列第一個數爲1，第二個數等於多少呢？然後發現，答案序列就是全都已知了，所以就是枚舉首位數就行了。
代碼也非常之少，要做的就是枚舉首位數字，然後查重。
代碼：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>
#include <queue>
#include <stack>
#include <map>
//鬼畜頭文件
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
typedef unsigned long long ULL;
typedef long long LL;
//鬼畜define
bool tong[1010];
int all[1010];
int n;
int main()
{
scanf("%d",&n);
for(int time=0;time<n-1;time++)scanf("%d",&all[time]);
int ans;
for(int time=1;time<all[0];time++)
{//枚舉起點的值
memset(tong,false,sizeof(tong));
int last=time;
tong[last]=true;
int flag=1;
for(int time1=0;time1<n-1;time1++)
{
last=all[time1]-last;
if(last<=0||last>n||tong[last]==true){flag=0;break;}
tong[last]=true;
}
if(flag){ans=time;break;}
}
printf("%d",ans);
for(int time=0;time<n-1;time++)
{
ans=all[time]-ans;
printf(" %d",ans);
}
printf("\n");
return 0;
}