D. Petya and Array 树状数组

题意：

给出一个数组，元素有正有负有0，问其区间和小于 t 的子区间的个数。

sum[ r ]-sum[ l-1 ]<t，其中sum是a的前缀和。

实现的方法就是从前往后对于每一个sum[ i ]，看在它前面有多少个大于等于sum[ i ] - t 的前缀和。

树状数组维护的是 i 前面有几个数小于等于它  

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <bits/stdc++.h>
#define  pi acos(-1.0)
#define  eps 1e-9
#define  fi first
#define  se second
#define  rtl   rt<<1
#define  rtr   rt<<1|1
#define  bug         printf("******
")
#define  mem(a,b)    memset(a,b,sizeof(a))
#define  name2str(x) #x
#define  fuck(x)     cout<<#x" = "<<x<<endl
#define  f(a)        a*a
#define  sf(n)       scanf("%d", &n)
#define  sff(a,b)    scanf("%d %d", &a, &b)
#define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  pf          printf
#define  FRE(i,a,b)  for(i = a; i <= b; i++)
#define  FREE(i,a,b) for(i = a; i >= b; i--)
#define  FRL(i,a,b)  for(i = a; i < b; i++)+
#define  FRLL(i,a,b) for(i = a; i > b; i--)
#define  FIN         freopen("data.txt","r",stdin)
#define  gcd(a,b)    __gcd(a,b)
#define  lowbit(x)   x&-x
#define rep(i,a,b) for(int i=a;i<b;++i)
#define per(i,a,b) for(int i=a-1;i>=b;--i)
using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 2e5 + 7;
int n;
LL a[maxn], sum[maxn], t, cnt[maxn], c[maxn];
void update ( int pos, LL x ) {
    while ( pos <= n + 1 ) {
        c[pos] += x;
        pos += lowbit ( pos );
    }
}
LL getsum ( int pos ) {
    LL sum = 0;
    while ( pos > 0 ) {
        sum += c[pos];
        pos -= lowbit ( pos );
    }
    return sum;
}
int main() {
    scanf ( "%d%lld", &n, &t );
    for ( int i = 1 ; i <= n ; i++ ) {
        scanf ( "%lld", &a[i] );
        sum[i] = sum[i - 1] + a[i];
        cnt[i] = sum[i];
    }
    sort ( cnt, cnt + n + 1 );
    LL ans = 0;
    for ( int i = 1 ; i <= n ; i++ ) {
        int  pos = lower_bound ( cnt, cnt + n + 1, sum[i - 1] ) - cnt + 1;//树状数组下标不能是0 所以需要将下标++
        update ( pos, 1 );
        pos = lower_bound ( cnt, cnt + n + 1, sum[i] - t + 1 ) - cnt; //求有多少个数小于等于sum[i] - t
        ans += ( i - getsum ( pos ) );//区间的总数是 i-1 小于等于的个数是  getsum ( pos ) - 1
    }
    printf ( "%lld
", ans );
    return 0 ;
}