CodeForces 21 C Stripe 2

Description

Once Bob took a paper stripe of n squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into three pieces so that the sum of numbers from
each piece is equal to the sum of numbers from any other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?

Input

The first input line contains integer n (1 ≤ n ≤ 105) — amount of squares in the stripe. The
second line contains n space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.

Output

Output the amount of ways to cut the stripe into three non-empty pieces so that the sum of numbers from each piece is equal to the sum of numbers from any other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.

Sample Input

Input

4
1 2 3 3

Output

1

Input

5
1 2 3 4 5

Output

0

把一个数组分三段，每段和都一样，问有几种分法，直接处理一下前缀和然后倒着计算一下即可。

#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<bitset>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int maxn = 1e5 + 10;
int T, n, m, a[maxn];
int f[maxn], sum;
long long ans;

int main()
{
while (~scanf("%d", &n))
{
f[0] = ans = sum = 0;
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
sum += a[i];
}
int now = 0;
for (int i = 1; i <= n; i++)
{
now += a[i];
f[i] = f[i - 1] + (now * 3 == sum);
}
now = 0;
for (int i = n; i > 2; i--)
{
now += a[i];
if (now * 3 == sum) ans += f[i - 2];
}
cout << ans << endl;
}
return 0;
}