题意：

询问 下标满足 a + b * k 的和是多少。

题解：

将询问分块。

将b >= blo直接算出答案。

否则存下来。

存下来之后，对于每个b扫一遍数组，然后同时处理相同b的询问。

 

代码：

#include
      
       using namespace std;#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);#define LL long long#define ULL unsigned LL#define fi first#define se second#define pb push_back#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define max3(a,b,c) max(a,max(b,c))#define min3(a,b,c) min(a,min(b,c))#define Show(x) cout << x << ' ';typedef pair
       
         pll;const int INF = 0x3f3f3f3f;const LL mod =  (int)1e9+7;const int N = 3e5 + 100;int n, m, k, p;int w[N];LL tot[N];LL ans[N];struct Node{    int a, b, id;}q[N];bool cmp(Node x1, Node x2){    return x1.b < x2.b;}int main(){    scanf("%d", &n);    for(int i = 1; i <= n; i++) scanf("%d", &w[i]);    scanf("%d", &p);    k = sqrt(n);    int t = 0, a, b;    LL tmp;    for(int i = 1; i <= p; i++){        scanf("%d%d", &a, &b);        if(b >= k){            tmp = 0;            for(int i = a; i <= n; i += b)                tmp += w[i];            ans[i] = tmp;        }        else {            q[t].a = a;            q[t].b = b;            q[t].id = i;            t++;        }    }    sort(q,q+t,cmp);    for(int i = 0; i < t; i++){        if(i == 0 || q[i].b != q[i-1].b){            b = q[i].b;            for(int i = n; i >= 1; i--){                if(i+b > n) tot[i] = w[i];                else tot[i] = tot[i+b] + w[i];            }        }        ans[q[i].id] = tot[q[i].a];    }    for(int i = 1; i <= p; i++){        printf("%I64d\n", ans[i]);    }    return 0;}