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Codeforces 712A Memory and Crow

note tput main family input ima sequence jpg ali


【題目鏈接】 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
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.

題意:

存在n個數b1,,b2,...,bn,

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

現給你a1,a2,...,an這n個數,問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

即數組b的第 i 項是數組a的第 i 項和第 i+1 項之和。

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【時間復雜度】 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 } 

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Codeforces 712A Memory and Crow