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  

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <set>
 7 #include <iostream>
 8 #include <map>
 9 #include <stack>
10 #include <string>
11 #include <vector>
12 #include <bits/stdc++.h>
13 #define  pi acos(-1.0)
14 #define  eps 1e-9
15 #define  fi first
16 #define  se second
17 #define  rtl   rt<<1
18 #define  rtr   rt<<1|1
19 #define  bug         printf("******\n")
20 #define  mem(a,b)    memset(a,b,sizeof(a))
21 #define  name2str(x) #x
22 #define  fuck(x)     cout<<#x" = "<<x<<endl
23 #define  f(a)        a*a
24 #define  sf(n)       scanf("%d", &n)
25 #define  sff(a,b)    scanf("%d %d", &a, &b)
26 #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
27 #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
28 #define  pf          printf
29 #define  FRE(i,a,b)  for(i = a; i <= b; i++)
30 #define  FREE(i,a,b) for(i = a; i >= b; i--)
31 #define  FRL(i,a,b)  for(i = a; i < b; i++)+
32 #define  FRLL(i,a,b) for(i = a; i > b; i--)
33 #define  FIN         freopen("data.txt","r",stdin)
34 #define  gcd(a,b)    __gcd(a,b)
35 #define  lowbit(x)   x&-x
36 #define rep(i,a,b) for(int i=a;i<b;++i)
37 #define per(i,a,b) for(int i=a-1;i>=b;--i)
38 using namespace std;
39 typedef long long  LL;
40 typedef unsigned long long ULL;
41 const int INF = 0x3f3f3f3f;
42 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
43 const int maxn = 2e5 + 7;
44 int n;
45 LL a[maxn], sum[maxn], t, cnt[maxn], c[maxn];
46 void update ( int pos, LL x ) {
47     while ( pos <= n + 1 ) {
48         c[pos] += x;
49         pos += lowbit ( pos );
50     }
51 }
52 LL getsum ( int pos ) {
53     LL sum = 0;
54     while ( pos > 0 ) {
55         sum += c[pos];
56         pos -= lowbit ( pos );
57     }
58     return sum;
59 }
60 int main() {
61     scanf ( "%d%lld", &n, &t );
62     for ( int i = 1 ; i <= n ; i++ ) {
63         scanf ( "%lld", &a[i] );
64         sum[i] = sum[i - 1] + a[i];
65         cnt[i] = sum[i];
66     }
67     sort ( cnt, cnt + n + 1 );
68     LL ans = 0;
69     for ( int i = 1 ; i <= n ; i++ ) {
70         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++
71         update ( pos, 1 );
72         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
73         ans += ( i - getsum ( pos ) );//区间的总数是 i-1 小于等于的个数是  getsum ( pos ) - 1
74     }
75     printf ( "%lld\n", ans );
76     return 0 ;
77 }

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