标签：amp printf code nts cts cond bubuko 题意 you
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:
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?
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.
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.
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 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 }
Codeforces 712A Memory and Crow
标签：amp printf code nts cts cond bubuko 题意 you
原文地址：https://www.cnblogs.com/fangxiaoqi/p/10193391.html
