Codeforces 712A Memory and Crow

Posted 2021-01-30 fangxiaoqi

【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, ..., b_n written in a row. For all i from 1 to n, values a_i are defined by the crows performing the following procedure:

• The crow sets a_i initially 0.
• The crow then adds b_i to a_i, subtracts b_i + 1, adds the b_i + 2 number, and so on until the n‘th number. Thus, a_i = b_i - b_i + 1 + b_i + 2 - b_i + 3....

Memory gives you the values a₁, a₂, ..., a_n, and he now wants you to find the initial numbers b₁, b₂, ..., b_n 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 a_i ( - 10⁹ ≤ a_i ≤ 10⁹) — the value of the i‘th number.

Output

Print n integers corresponding to the sequence b₁, b₂, ..., b_n. 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.

 

 

 

 

题意：

存在n个数b1,，b2，...，bn，

a1，a2， ...， an 是通过等式 ai = bi - b(i+1) + b(i+2) - b(i+3)....(±)bn 得到的。

现给你a1，a2，...，an这n个数，问b1， b2， ...， bn是多少？

【分析】

∵ ai = bi - b(i+1) + b(i+2) - b(i+3)....

∴ bi = ai + b(i+1) - b(i+2) +b(i+3)....

又 ∵ b(i+1) = a(i+1) + b(i+2) - b(i+3) +b(i+4)....

∴ bi = ai + [ a(i+1) + b(i+2) - b(i+3) +b(i+4).... ] - b(i+2) +b(i+3)....

        = ai + a(i+1)

而且，an = bn

即数组b的第 i 项是数组a的第 i 项和第 i+1 项之和。

【时间复杂度】 O(n)

---

 

 1 #include<stdio.h>
 2 int main(){
 3     int n,a,b,i;
 4     while(1){
 5         scanf("%d",&n);
 6         for(i = 1;i <= n; i++){
 7             scanf("%d",&a);
 8             if(i>1)
 9                 printf("%d ",a+b);
10             b = a;
11         }
12         printf("%d",b);
13     }
14     return 0;
15 }