BZOJ4627 [BeiJing2016]回转寿司_nor前缀

求前缀和，相当于对于每个i求1<=j<i，L<=s[i]-s[j]<=R的个数然后加起来，用权值线段树搞就好了

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<ctime>
#include<vector>
#include<stack>
#include<set>
#include<bitset>
#include<map>
#include<queue>
using namespace std;
#define MAXN 100010
#define MAXM 8000010
#define MOD 1000000007
#define INF 10000000000ll
#define eps 1e-8
#define ll long long
int n,L,R;
int siz[MAXM],son[MAXM][2];
int tot;
int rt;
void change(int &x,ll y,ll z,ll p){
	if(!x){
		x=++tot;
	}
	siz[x]++;
	if(y==z){
		return ;
	}
	ll mid=y+z>>1;
	if(p<=mid){
		change(son[x][0],y,mid,p);
	}else if(p>mid){
		change(son[x][1],mid+1,z,p);
	}
}
int ask(int x,ll y,ll z,ll l,ll r){
	if(!x){
		return 0;
	}
	if(y==l&&z==r){
		return siz[x];
	}
	ll mid=y+z>>1;
	if(r<=mid){
		return ask(son[x][0],y,mid,l,r);
	}else if(l>mid){
		return ask(son[x][1],mid+1,z,l,r);
	}else{
		return ask(son[x][0],y,mid,l,mid)+ask(son[x][1],mid+1,z,mid+1,r);
	}
}
ll ans;
int main(){
	int i,x;
	ll s=0;
	scanf("%d%d%d",&n,&L,&R);
	for(i=1;i<=n;i++){
		scanf("%d",&x);
		change(rt,-INF,INF,s);
		s+=x;
		ans+=ask(rt,-INF,INF,max(-INF,s-R),min(INF,s-L));
	}
	printf("%lld\n",ans);
	return 0;
}
 
/*
*/