CodeForces 103D Time to Raid Cowavans 询问分块

Time to Raid Cowavans　　

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<bits/stdc++.h>
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<int,int> pll;
const int INF = 0x3f3f3f3f;
const LL mod =  (int)1e9+;
const int N = 3e5 + ;
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 = ; i <= n; i++) scanf("%d", &w[i]);
    scanf("%d", &p);
    k = sqrt(n);
    int t = , a, b;
    LL tmp;
    for(int i = ; i <= p; i++){
        scanf("%d%d", &a, &b);
        if(b >= k){
            tmp = ;
            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 = ; i < t; i++){
        if(i ==  || q[i].b != q[i-].b){
            b = q[i].b;
            for(int i = n; i >= ; 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 = ; i <= p; i++){
        printf("%I64d\n", ans[i]);
    }
    return ;
}