【CF103D】Time to Raid Cowavans（分块）

题意：

思路：院赛防AK题，然而还没来得及做就被数据出锅的题坑了……

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 #include<bitset>
 #include<ctime>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N      310000
 #define M      51
 #define MOD 1000000007
 #define eps 1e-8
 #define pi     acos(-1)
 #define oo     1e9

 ll a[N],s[N],ans[N];
 int n;

 struct node
 {
     int x,y,id;
 }b[N];

 bool cmp(node a,node b)
 {
     return a.y<b.y;
 }

 int main()
 {
     int n;
     scanf("%d",&n);
     for(int i=;i<=n;i++) scanf("%lld",&a[i]);
     int q;
     scanf("%d",&q);
     for(int i=;i<=q;i++)
     {
         scanf("%d%d",&b[i].x,&b[i].y);
         b[i].id=i;
     }
      sort(b+,b+q+,cmp);
     for(int i=;i<=q;i++)
      if(b[i].y<=(int)sqrt(n))
      {
           if(i>&&b[i].y==b[i-].y) ans[b[i].id]=s[b[i].x];
            else
            {
                for(int j=;j<=n;j++) s[j]=;
                for(int j=n;j>=;j--)
                {
                    s[j]=a[j];
                    if(j+b[i].y<=n) s[j]+=s[j+b[i].y];
             }
             ans[b[i].id]=s[b[i].x];
            }
      }
       else
       {
             ll t=;
             int j=b[i].x;
             while(j<=n)
             {
                 t+=a[j];
                 j+=b[i].y;
             }
             ans[b[i].id]=t;
       }
     for(int i=;i<=q;i++) printf("%lld\n",ans[i]);
      return ;
 }