Codeforces Round #370 (Div. 2) A. Memory and Crow 水题

A. Memory and Crow

题目连接：

http://codeforces.com/contest/712/problem/A

Description

There are n integers b1, b2, ..., bn written in a row. For all i from 1 to n, values ai are defined by the crows performing the following procedure:

The crow sets ai initially 0.
The crow then adds bi to ai, subtracts bi + 1, adds the bi + 2 number, and so on until the n'th number. Thus, ai = bi - bi + 1 + bi + 2 - bi + 3.... 

Memory gives you the values a1, a2, ..., an, and he now wants you to find the initial numbers b1, b2, ..., bn written in the row? Can you do it?

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of integers written in the row.

The next line contains n, the i'th of which is ai ( - 109 ≤ ai ≤ 109) — the value of the i'th number.

Output

Print n integers corresponding to the sequence b1, b2, ..., bn. It's guaranteed that the answer is unique and fits in 32-bit integer type.

Sample Input

5
6 -4 8 -2 3

Sample Output

2 4 6 1 3

Hint

题意

题解：

显然知道an=bn,然后剩下的可以手动推一下公式，就可以递推出来了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int a[maxn],n;
long long b[maxn];
int main()
{
    scanf("%d",&n);
    long long sum = 0;
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=n;i>=1;i--)
    {
        sum*=-1;
        b[i]=a[i]-sum;
        sum+=b[i];
    }
    for(int i=1;i<=n;i++)
        printf("%lld ",b[i]);
    printf("\n");
}