Petya has an array $\mathrm{aa\; consisting\; of\; nn\; integers.\; He\; has\; learned\; partial\; sums\; recently,\; and\; now\; he\; can\; calculate\; the\; sum\; of\; elements\; on\; any\; segment\; of\; the\; array\; really\; fast.\; The\; segment\; is\; a\; non-empty\; sequence\; of\; elements\; standing\; one\; next\; to\; another\; in\; the\; array.}$

Now he wonders what is the number of segments in his array with the sum less than $\mathrm{tt.\; Help\; Petya\; to\; calculate\; this\; number.}$

More formally, you are required to calculate the number of pairs $\mathrm{l,rl,r\; (l\le rl\le r)\; such\; that\; al+al+1+?+ar?1+ar<tal+al+1+?+ar?1+ar<t.}$

The first line contains two integers

The second line contains a sequence of integers ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an\; (|ai|\le 109|ai|\le 109)\; —\; the\; description\; of\; Petya‘s\; array.\; Note\; that\; there\; might\; be\; negative,\; zero\; and\; positive\; elements.}}^{}$

Print the number of segments in Petya‘s array with the sum of elements less than

5 4
5 -1 3 4 -1

5

3 0
-1 2 -3

4

4 -1
-2 1 -2 3

3

In the first example the following segments have sum less than $\mathrm{44:}$

• $[2,2][2,2],\; sum\; of\; elements\; is\; ?1?1$
• $[2,3][2,3],\; sum\; of\; elements\; is\; 22$
• $[3,3][3,3],\; sum\; of\; elements\; is\; 33$
• $[4,5][4,5],\; sum\; of\; elements\; is\; 33$
• $[5,5][5,5],\; sum\; of\; elements\; is\; ?1?1$

先補上樹狀數組的解法

sum[i] - sum[j] < t , 1 <= j < i,所以sum[i] - t < sum[j]

因為j是i之前的前綴和，那麽我們可以找當前sum[j],小於等於sum[i]-t的，然後用ans+=（i-query），然後一直wa，看了網上題解，加入一個0的虛擬節點後（相當於每個當前前綴和），樹狀數組記錄當前時刻1到n+1,而不是1到n；

這樣就可以知道當前前綴和之前的前綴和出現情況。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long ll;
 5 const int maxn = 2e5+5;
 6 
 7 ll tree[maxn];
 8 ll sum[maxn];
 9 ll num[maxn];
10 int n;
11 ll t;
12 
13 int lowbit(int x)
14 {
15     return x&(-x);
16 }
17 
18 void add(int x)
19 {
20     for(int i=x;i<=n+1;i+=lowbit(i))
21     {
22         tree[i]++;
23     }
24 }
25 
26 ll query(int x)
27 {
28     ll ans = 0;
29     for(int i=x;i>0;i-=lowbit(i))
30     {
31         ans += tree[i];
32     }
33     return ans;
34 }
35 
36 int main()
37 {
38      scanf("%d%lld",&n,&t);
39      for(int i=1;i<=n;i++)
40      {
41          scanf("%lld",&sum[i]);
42          sum[i] += sum[i-1];
43          num[i] = sum[i];
44      }
45      sort(num,num+n+1);
46      ll ans = 0;
47      for(int i=1;i<=n;i++)
48      {
49          add(lower_bound(num,num+n+1,sum[i-1])-num+1);
50          ans += i - query(upper_bound(num,num+n+1,sum[i]-t)-num);
51      }
52     printf("%lld\n",ans);
53 }

Petya and Array CodeForces - 1042D （樹狀數組）