小纪的作业题

题目描述

莫队算法

他符合莫队算法的特征！
于是用g[x]表示x这类数对答案的贡献，然后用num[x]记录x出现的次数即可。预处理一下每个数的逆元。注意num[x]<=0时g[x]应当为0。

参考程序

#include<cstdio>
#include<cmath>
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const ll maxn=100000+10,mo=1000000007;
struct dong{
    ll l,r,id;
};
dong ask[maxn];
ll ans[maxn],belong[maxn],a[maxn];
ll g[maxn],num[maxn],ni[maxn];
ll i,j,k,l,r,t,n,m,c,now;
bool cmp(dong a,dong b){
    if (belong[a.l]<belong[b.l]) return 1;
    else if (belong[a.l]==belong[b.l]&&a.r<b.r) return 1;
    else return 0;
}
ll quicksortmi(ll x,ll y){
    if (!y) return 1;
    ll t=quicksortmi(x,y/2);
    t=t*t%mo;
    if (y%2) t=t*x%mo;
    return t;
}
void out(ll x){
    now=((now-g[x])%mo+mo)%mo;
    num[x]--;
    if (num[x]==0) g[x]=0;else g[x]=g[x]*ni[x]%mo;
    now=(now+g[x])%mo;
}
void in(ll x){
    now=((now-g[x])%mo+mo)%mo;
    num[x]++;
    if (num[x]==1) g[x]=x;else g[x]=(g[x]*x)%mo;
    now=(now+g[x])%mo;
}
int main(){
    scanf("%lld%lld",&n,&m);
    fo(i,1,n) scanf("%lld",&a[i]);
    c=floor(sqrt(n))+1;
    fo(i,1,n) belong[i]=(i-1)/c+1;
    fo(i,1,m) scanf("%lld%lld",&ask[i].l,&ask[i].r),ask[i].id=i;
    sort(ask+1,ask+m+1,cmp);
    fo(i,1,100000) ni[i]=quicksortmi(i,mo-2);
    l=r=1;
    num[a[1]]=1;
    g[a[1]]=a[1];
    now=g[a[1]];
    fo(i,1,m){
        while (l<ask[i].l){
            out(a[l]);
            l++;
        }
        while (l>ask[i].l){
            l--;
            in(a[l]);
        }
        while (r<ask[i].r){
            r++;
            in(a[r]);
        }
        while (r>ask[i].r){
            out(a[r]);
            r--;
        }
        ans[ask[i].id]=now;
    }
    fo(i,1,m) printf("%lld\n",ans[i]);
}