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.

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
Note
In the first sample test, the crows report the numbers 6,?-?4, 8,?-?2, and 3 when he starts at indices 1, 2, 3, 4 and 5 respectively. It is easy to check that the sequence 2 4 6 1 3 satisfies the reports. For example, 6?=?2?-?4?+?6?-?1?+?3, and ?-?4?=?4?-?6?+?1?-?3.

In the second sample test, the sequence 1, ?-?3, 4, 11, 6 satisfies the reports. For example, 5?=?11?-?6 and 6?=?6.

分析可得bi=ai+ai+1

#include <stdio.h> #include <stdlib.h>/* run this program using the console pauser or add your own getch, system("pause") or input loop */int main(int argc, char *argv[]) {int n;scanf("%d",&n);int i;int a[100000];for(i=0;i<n;i++){scanf("%d",&a[i]);}for(i=0;i<n-1;i++){printf("%d ",a[i]+a[i+1]);}printf("%d\n",a[i]);return 0; }

總結

以上是生活随笔為你收集整理的Memory and Crow的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。