CodeForces 103 D Time to Raid Cowavans

Time to Raid Cowavans

题意：一共有n头牛， 每头牛有一个重量，m次询问， 每次询问有a，b 求出 a，a+b，a+2b的牛的重量和。

题解：对于m次询问，b>sqrt(n)的时候我们直接把结果跑出来，当b<sqrt(n)的时候我们离线询问，算出所有一个b的任意一起点的值。 复杂度为 q*sqrt(n);

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 ;
 }

CF103 D