CRAN02 - Roommate Agreement
Leonard was always sickened by how Sheldon considered himself better than him. To decide once and for all who is better among them they decided to ask each other a puzzle. Sheldon pointed out that according to Roommate Agreement Sheldon will ask first. Leonard seeing an opportunity decided that the winner will get to rewrite the Roommate Agreement.
Sheldon thought for a moment then agreed to the terms thinking that Leonard will never be able to answer right. For Leonard, Sheldon thought of a puzzle which is as follows. He gave Leonard n numbers, which can be both positive and negative. Leonard had to find the number of continuous sequence of numbers such that their sum is zero.
For example if the sequence is- 5, 2, -2, 5, -5, 9
There are 3 such sequences
2, -2
5, -5
2, -2, 5, -5
Since this is a golden opportunity for Leonard to rewrite the Roommate Agreement and get rid of Sheldon‘s ridiculous clauses, he can‘t afford to lose. So he turns to you for help. Don‘t let him down.
Input
First line contains T - number of test cases
Second line contains n - the number of elements in a particular test case.
Next line contain n elements, ai (1<=i<= n) separated by spaces.
Output
The number of such sequences whose sum if zero.
Constraints
1<=t<=5
1<=n<=10^6
-10<= ai <= 10
Example
Input:
2
4
0 1 -1 0
6
5 2 -2 5 -5 9
Output:
63
题意
给你一个序列，里面n(10^6)个数字，问这些数字相加为0的区间有多少个
思路
看样例解释我们可以知道，可以利用前缀和来计算，a[i]=a[i]+a[i-1]这样，然后用map来存a[i]出现的次数，
1 /*
2 Name: hello world.cpp
3 Author: AA
4 Description: 唯代码与你不可辜负
5 */
6 #include<bits/stdc++.h>
7 using namespace std;
8 #define LL long long
9 int main() {
10 int t;
11 cin >> t;
12 while(t--) {
13 int n;
14 cin >> n;
15 LL a[n];
16 map<LL, LL> cnt;
17 cin >> a[0];
18 cnt[a[0]]++;
19 for(int i = 1; i < n; i++) {
20 cin >> a[i];
21 a[i] += a[i - 1];
22 cnt[a[i]]++;
23 }
24 map<LL, LL>::iterator it;
25 LL ans = 0;
26 for(it = cnt.begin(); it != cnt.end(); it++) {
27 if(it->first == 0)
28 ans += it->second + it->second * (it->second - 1) / 2;
29 else
30 ans += it->second * (it->second - 1) / 2;
31 }
32 cout << ans << endl;
33 }
34 return 0;
35 }