nbut 1457 Sona
https://ac.2333.moe/Problem/view.xhtml?id=1457 题意:求一段区间内数字个数的立方和。 思路:由于一共才1E5,而数字1E9,所以先离散化,再莫队,类似小z的袜子。 注意 :%lld会WA,要用%I64d
/* ***********************************************
Author :111qqz
Created Time :2016年02月17日 星期三 16时11分00秒
File Name :code/nbut/1457.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=1E5+7;
7int a[N],b[N];
8LL cnt[N];
9int pos[N];
10LL ans[N];
11LL sum;
12int n,m;
1struct node
2{
3 int l,r;
4 int id;
1 bool operator < (node b)const
2 {
3 if (pos[l]==pos[b.l]) return r<b.r;
4 return pos[l]<pos[b.l];
5 }
6}q[N];
void update(int x,int d)
{
1 sum-=cnt[a[x]]*cnt[a[x]]*cnt[a[x]];
2 cnt[a[x]]+=d;
3 sum +=cnt[a[x]]*cnt[a[x]]*cnt[a[x]];
1}
2int main()
3{
4 #ifndef ONLINE_JUDGE
5 freopen("code/in.txt","r",stdin);
6 #endif
1 while (scanf("%d",&n)!=EOF)
2 {
3 ms(cnt,0);
4 ms(ans,0);
5 sum = 0LL;
1 int siz = 330;//sqrt(100000);
2 for ( int i = 1 ; i <= n ; i++)
3 {
4 scanf("%d",&b[i]);
5 pos[i] = (i-1)/siz;
6 a[i] = b[i];
7 }
1 sort(b+1,b+n+1); //离散化
2 int t = unique(b+1,b+n+1)-b-1;
3 for ( int i = 1 ; i <= n ; i++) a[i]=lower_bound(b+1,b+t+1,a[i])-b;
1 scanf("%d",&m);
2 for ( int i = 1 ; i <= m ; i++)
3 {
4 scanf("%d %d",&q[i].l,&q[i].r);
5 q[i].id = i ;
6 }
7 sort(q+1,q+m+1);
1 int pl=1,pr=0;
2 int id,l,r;
3 for ( int i = 1 ; i <= m ; i++)
4 {
5 id = q[i].id;
6 r = q[i].r;
7 l = q[i].l;
1 if (pr<r)
2 {
3 for ( int j = pr +1 ; j <= r ; j++) update(j,1);
4 }
5 else
6 {
7 for (int j = r+1 ; j <= pr ; j++) update(j,-1);
8 }
9 pr = r;
1 if (l<pl)
2 {
3 for ( int j = l ; j <= pl-1 ; j++) update(j,1);
4 }
5 else
6 {
7 for (int j = pl ; j <= l-1 ; j++) update(j,-1);
8 }
9 pl = l;
ans[id] = sum;
}
for ( int i = 1 ; i <= m ; i++) printf("%I64d\n",ans[i]); //用%lld会WA...也不给个警告
}
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}