NBUT 1457 Sona [简单莫队]_- sona nbut

题意: 询问M次区间不同数各自出现次数的立方和。

解法：简单莫队

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")


#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define drep(i,a,b) for(int i=a;i>=b;i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef __int64 ll;

const int NN = 200200;

int n,qn;
int a[NN],b[NN],bn;
int block[NN],bnum;
ll cot[NN],ans[NN],sum;

struct query{
    int l,r,idx;
}q[NN];
bool cmp(query x,query y){
    if(block[x.l]==block[y.l])return x.r<y.r;
    return block[x.l]<block[y.l];
}
void init(){
    sum=0;
    bn=n;bnum=max(1,int(sqrt(double(n))));
    rep(i,1,n)cot[i]=0;
    rep(i,1,n)block[i]=(i-1)/bnum+1;
    rep(i,1,n)scanf("%d",&a[i]),b[i]=a[i];
    sort(b+1,b+1+bn);
    bn=unique(b+1,b+1+bn)-b-1;
    rep(i,1,n)a[i]=findx(a[i]);
    scanf("%d",&qn);
    rep(i,1,qn)scanf("%d%d",&q[i].l,&q[i].r),q[i].idx=i;
    sort(q+1,q+1+qn,cmp);
}
void update(int pos,bool add){
    sum-=cot[pos]*cot[pos]*cot[pos];
    if(add)cot[pos]++;
    else cot[pos]--;
    sum+=cot[pos]*cot[pos]*cot[pos];
}
void solve(){
    int l=1,r=0;
    rep(i,1,qn){
        for(;r<q[i].r;r++)update(a[r+1],1);
        for(;r>q[i].r;r--)update(a[r],0);
        for(;l<q[i].l;l++)update(a[l],0);
        for(;l>q[i].l;l--)update(a[l-1],1);
        ans[q[i].idx]=sum;
    }
    rep(i,1,qn)printf("%I64d\n",ans[i]);
}
int main(){
    while(scanf("%d",&n)!=EOF){
        init();
        solve();
    }
}