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：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
 4 #define LL long long
 5 #define ULL unsigned LL
 6 #define fi first
 7 #define se second
 8 #define pb push_back
 9 #define lson l,m,rt<<1
10 #define rson m+1,r,rt<<1|1
11 #define max3(a,b,c) max(a,max(b,c))
12 #define min3(a,b,c) min(a,min(b,c))
13 #define Show(x) cout << x << ‘ ‘;
14 typedef pair<int,int> pll;
15 const int INF = 0x3f3f3f3f;
16 const LL mod =  (int)1e9+7;
17 const int N = 3e5 + 100;
18 int n, m, k, p;
19 int w[N];
20 LL tot[N];
21 LL ans[N];
22 struct Node{
23     int a, b, id;
24 }q[N];
25 bool cmp(Node x1, Node x2){
26     return x1.b < x2.b;
27 }
28 int main(){
29     scanf("%d", &n);
30     for(int i = 1; i <= n; i++) scanf("%d", &w[i]);
31     scanf("%d", &p);
32     k = sqrt(n);
33     int t = 0, a, b;
34     LL tmp;
35     for(int i = 1; i <= p; i++){
36         scanf("%d%d", &a, &b);
37         if(b >= k){
38             tmp = 0;
39             for(int i = a; i <= n; i += b)
40                 tmp += w[i];
41             ans[i] = tmp;
42         }
43         else {
44             q[t].a = a;
45             q[t].b = b;
46             q[t].id = i;
47             t++;
48         }
49     }
50     sort(q,q+t,cmp);
51     for(int i = 0; i < t; i++){
52         if(i == 0 || q[i].b != q[i-1].b){
53             b = q[i].b;
54             for(int i = n; i >= 1; i--){
55                 if(i+b > n) tot[i] = w[i];
56                 else tot[i] = tot[i+b] + w[i];
57             }
58         }
59         ans[q[i].id] = tot[q[i].a];
60     }
61     for(int i = 1; i <= p; i++){
62         printf("%I64d\n", ans[i]);
63     }
64     return 0;
65 }

CF103 D

原文地址：https://www.cnblogs.com/MingSD/p/9104959.html