CF 466C Number of Ways

C. Number of Ways

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.

More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that .

Input

The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1],a[2], ..., a[n] (|a[i]| ≤  109) — the elements of array a.

Output

Print a single integer — the number of ways to split the array into three parts with the same sum.

Sample test(s)

input

51 2 3 0 3

output

2

input

40 1 -1 0

output

1

input

24 1

output

0

是一个技巧题，s[i] = (s[n-1]-s[i])/2,这个关系可以推出3*s[i] == s[n-1].这样子能找出满足条件的一头一尾，但是中间还有一个点。每遇到一个能位于中间的点，就把CNT加上，CNT可以看成是前面满足条件的第一个点的个数。

#include<iostream>using namespace std;long long num[500005]={0};int main(){int n;cin >> n;for(int i = 0;i<n;i++){cin >> num[i];if(i>0)num[i] = num[i]+num[i-1];}long long ans = 0;int cnt=0;for(int i=0;i<n-1;i++){if(3*num[i] == num[n-1]*2)ans+=cnt;if(3*num[i] == num[n-1]){cnt++;}}cout<<ans<<endl;return 0;} 

然而实际上。题意是说，将一串的序列分成三部分，那么假设第一部分总和是a,前两部分总和就是2*a,整个总和就是3*a.所以直接将num[n-1]/3。可以数出a和2*a的个数。

#include<iostream>using namespace std;long long num[500005]={0};int main(){int n;cin >> n;for(int i = 0;i<n;i++){cin >> num[i];if(i>0)num[i] = num[i]+num[i-1];}long long ans = 0;int cnt=0;if(num[n-1]%3==0){for(int i=0;i<n-1;i++){if(num[i] == num[n-1]/3*2)ans+=cnt;if(num[i]==num[n-1]/3){cnt++;}}}cout<<ans<<endl;return 0;}