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.
Examples
input
5
1 2 3 0 3 output
2 input
4
0 1 -1 0 output
1 input
2
4 1 output
0
題目的意思為輸入一個數n，然後輸入n個數，計算有幾種方法讓這n個數分為和相對的三份。
n的數字為十萬級，需要定義全局變量才能運作，而且n需要定義為long long型，求解思路為：設定一個記錄和的數組，然後将其中和的1/3和2/3的資料記錄在另一個數組fenlie中，設定一個權值k=0，遇到1/3，k++，遇到2/3，sumn+=k。 #include<iostream>
#define N 500005
using namespace std;
long long sum[N];
long long number[N];
long long fenlie[N];
int main()
{
long long i, n, k=0;
long long sumn = 0;
//int *sum = new int[N];
//int *number = new int[N];
//int *fenlie = new int[N];
sum[0] = 0;
cin >> n;
for (i = 1;i <= n;i++)
{
cin >> number[i];
sum[i] += sum[i - 1] + number[i];
}
if (sum[n] % 3 != 0)
cout << '0' << endl;
else
{
long long temp = sum[n] / 3;
long long j = 0;
for (i = 1;i <= n;i++)
{
if (sum[i] == temp || sum[i] == 2 * temp)
{
fenlie[j] = sum[i];
j++;
}
}
if (temp == 0)
sumn = (j - 1)*(j - 2) / 2;
else
{
for (i = 0;i < j;i++)
{
if (fenlie[i] == temp)
k++;
else if (fenlie[i] == 2 * temp)
sumn += k;
}
}
cout << sumn << endl;
}
/*delete sum;
delete fenlie;
delete number;*/
return 0;
}