ZCMU-1036

1036: Shepherd

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 140  Solved: 29

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Description

Hehe keeps a flock of sheep, numbered from 1 to n and each with a weight wi. To keep the sheep healthy, he prepared some training for his sheep. Everytime he selects a pair of numbers (a,b), and chooses the sheep with
number a, a+b, a+2b, … to get trained. For the distance between the sheepfold and the training site is too far, he needs to arrange a truck with appropriate loading capability to transport those sheep. So he wants to know the total weight of the
sheep he selected each time, and he finds you to help him.

Input

There’re several test cases. For each case:

The first line contains a positive integer n (1≤n≤10^5)---the number of sheep Hehe keeps.

The second line contains n positive integer wi(1≤n≤10^9), separated by spaces, where the i-th number describes the weight of the i-th sheep.

The third line contains a positive integer q (1≤q≤10^5)---the number of training plans Hehe prepared.

Each following line contains integer parameters a and b (1≤a,b≤n)of the corresponding plan.

Output

For each plan (the same order in the input), print the total weight of sheep selected.

Sample Input

5

1 2 3 4 5

3

1 1

2 2

3 3

Sample Output

15

6

3

【解析】

这道题的大致意思就是给你个数组，然后三次查找，给你选定范围比如说1 1则是从1开始每隔1个数，这道题的话我超时了三次，最后是在两地方做了改动一个就是算了数组的总和，另一个就是判断如果p=1并且q=1那就直接输出总和就好。

#include<iostream>
#include<string>
#include<vector>
#include<cstdio>
using namespace std;
vector<int>a;
int main()
{
int n,m,i,p,q,num,j;
long long sum1,sum2;
while(~scanf("%d",&n))
{
a.clear();
sum1=0;
for(i=0;i<n;i++)
{
scanf("%d",&num);
a.push_back(num);
sum1+=num;
}
scanf("%d",&m);
while(m--)
{
sum2=0;
scanf("%d%d",&p,&q);
if(p==1&&q==1)
{
printf("%lld\n",sum1);
continue;
}
for(j=p-1;j<n;j+=q)
sum2+=a[j];
printf("%lld\n",sum2);
}
}
return 0;
}