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("******\n")
 #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 + ;
 int n;
 LL a[maxn], sum[maxn], t, cnt[maxn], c[maxn];
 void update ( int pos, LL x ) {
     while ( pos <= n +  ) {
         c[pos] += x;
         pos += lowbit ( pos );
     }
 }
 LL getsum ( int pos ) {
     LL sum = ;
     while ( pos >  ) {
         sum += c[pos];
         pos -= lowbit ( pos );
     }
     return sum;
 }
 int main() {
     scanf ( "%d%lld", &n, &t );
     for ( int i =  ; i <= n ; i++ ) {
         scanf ( "%lld", &a[i] );
         sum[i] = sum[i - ] + a[i];
         cnt[i] = sum[i];
     }
     sort ( cnt, cnt + n +  );
     LL ans = ;
     for ( int i =  ; i <= n ; i++ ) {
         int  pos = lower_bound ( cnt, cnt + n + , sum[i - ] ) - cnt + ;//树状数组下标不能是0 所以需要将下标++
         update ( pos,  );
         pos = lower_bound ( cnt, cnt + n + , sum[i] - t +  ) - 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\n", ans );
     return  ;
 }