Codeforces 712A Memory and Crow

参考自：/ECJTUACM-873284962/p/6374955.html

【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 arenintegersb_{1, b2, ..., bnwritten in a row. For allifrom1 ton, valuesaiare defined by the crows performing the following procedure:}

The crow setsa_{iinitially0.The crow then addsbitoai, subtractsbi + 1, adds thebi + 2number, and so on until then'th number. Thus,ai = bi - bi + 1 + bi + 2 - bi + 3....}

Memory gives you the valuesa_{1, a2, ..., an, and he now wants you to find the initial numbersb1, b2, ..., bnwritten in the row? Can you do it?}

Input

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

The next line containsn, thei'th of which isa_{i( - 10^{9 ≤ ai ≤ 109)— the value of thei'th number.}}

Output

Printnintegers corresponding to the sequenceb_{1, 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

2 4 6 1 3

5

3 -2 -1 5 6

1 -3 4 11 6

In the first sample test, the crows report the numbers6, - 4,8, - 2, and3 when he starts at indices1,2,3,4 and5 respectively. It is easy to check that the sequence24613 satisfies the reports. For example,6 = 2 - 4 + 6 - 1 + 3, and - 4 = 4 - 6 + 1 - 3.

In the second sample test, the sequence1, - 3,4,11,6 satisfies the reports. For example,5 = 11 - 6 and6 = 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)

【代码实现】

1 #include<stdio.h> 2 int main(){ 3int n,a,b,i; 4while(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}14return 0;15 }