BZOJ4627 前缀和 + 权值线段树

https://www.lydsy.com/JudgeOnline/problem.php?id=4627

题意：求序列中和在L到R之间的字串种数。

要求的是和的范围，我们可以考虑先求一个前缀和pre,然后每个点j的贡献就是L <= pre[j] - pre[i] <= R（i < j)的i的种数了,移项一下变成

pre[j] - R <= pre[i] <= pre[j] - L

我们就可以考虑做个权值线段树维护一下所有pre,每次求贡献的时候做一个区间查询就可以了。

注意点:

1.由于范围太大又有负数，线段树要动态开点。

2.开局要加一个pre[0] = 0的前缀和

3.线段树里不必要开的点尽量不开否则可能由于空间WA

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d
", x)
#define Prl(x) printf("%lld
",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 1e5 + 10;
const LL INF = 1e10 + 2e9;
const int mod = 1e9 + 7; 
int N,M,K;
LL L,R;
struct Tree{
    LL sum;
    int lt,rt;
    void init(){lt = rt = sum = 0;}
}tree[maxn * 60];
int tot;
void check(int &t){
    if(t) return;
    t = ++tot;
    tree[t].init();
}
void Pushup(int &t){
    int ls = tree[t].lt,rs = tree[t].rt;
    tree[t].sum = tree[ls].sum + tree[rs].sum;
}
void update(int &t,LL l,LL r,LL x){
    check(t);
    if(l == r){
        tree[t].sum++;
        return;
    } 
    LL m = l + r >> 1;
    if(x <= m) update(tree[t].lt,l,m,x);
    else update(tree[t].rt,m + 1,r,x);
    Pushup(t);
}
LL query(int &t,LL l,LL r,LL ql,LL qr){
    if(!t) return 0;
    if(ql <= l && r <= qr) return tree[t].sum;
    LL m = l + r >> 1;
    if(qr <= m) return query(tree[t].lt,l,m,ql,qr);
    else if(ql > m) return query(tree[t].rt,m + 1,r,ql,qr);
    else return query(tree[t].lt,l,m,ql,m) + query(tree[t].rt,m + 1,r,m + 1,qr);
}
LL pre[maxn];
int main(){
    N = read(); L = read(); R = read(); // sum[j] - L >= sum[i - 1] >= sum[j] - R;
    for(int i = 1; i <= N; i ++) pre[i] = read() + pre[i - 1];
    LL ans = 0;
    int root = 0;
    update(root,-INF,INF,0);
    for(int i = 1; i <= N ; i ++){
        if(pre[i] - R <= pre[i] - L) ans += query(root,-INF,INF,pre[i] - R,pre[i] - L);
        update(root,-INF,INF,pre[i]);
    }
    Prl(ans);
    return 0;
}