题目链接：

A. Memory and Crow

time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

Examples
input
5
6 -4 8 -2 3
output
2 4 6 1 3
input
5
3 -2 -1 5 6
output
1 -3 4 11 6 

题意：

给了a，让求b；

思路：

水水水;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
  
using namespace std;
  
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
  
typedef  long long LL;
  
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}
  
const int mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=(1<<20)+10;
const int maxn=1e5+110;
const double eps=1e-12;
 

int n,a[maxn];
int main()
{
    read(n);
    For(i,1,n)read(a[i]);
    For(i,1,n)printf("%d ",a[i]+a[i+1]);

    return 0;
}