ZCMU 1036: Shepherd

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

【分析】题意是n只羊，从第c只开始，每隔b-1只羊，就计算重量和

题目简单，但有两个坑，如果把sum定义为int型会wa,不单独考虑a==1&&b==1时会Time Limit Exceed，我觉得是题目设置问题

#include<iostream>  
#include<cstdio>  
using namespace std;  
int main()  
{  
    int a[100000];  
    int n;  
    while(~scanf("%d",&n))  
    {  
        long long s1=0;  
        for(int i=0;i<n;i++)  
        {  
            scanf("%d",&a[i]);  
            s1+=a[i];  
        }  
        int pp;scanf("%d",&pp);  
        while(pp--)  
        {  
            long long sum=0;  
            int b,c;  
            scanf("%d%d",&c,&b);  
            if(c==1&&b==1){printf("%lld\n",s1);continue;}  
            for(int i=c-1;i<n;i+=b)sum+=a[i];  
            printf("%lld\n",sum);  
        }  
    }  
    return 0;  
}