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

好題，dp思維或者二分可解。但正解應該是O(n)

首先，我們用陣列lsum[]記錄字首和。此時，正常思想是迴圈i,j兩個位置變數，lsum[i]=sum/3，lsum[j]=sum*2/3，而這樣的做法是O(n)，顯然不可行。另一個方法就是，一遍遍歷找到lsum[i]=sum*2/3的位置，在找到這個位置之前，我們需要記錄lsum[i]=sum/3的個數，我們用一個變數儲存，這樣在找到lsum[i]=sum*2/3的位置時，我們就可以計算出總的情況數了。這樣的複雜度是O(n)。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<stack>
#include<queue>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#define eps 1e-8
#define zero(x) (((x>0?(x):-(x))-eps)
#define mem(a,b) memset(a,b,sizeof(a))
#define memmax(a) memset(a,0x3f,sizeof(a))
#define pfn printf("\n")
#define ll __int64
#define ull unsigned long long
#define sf(a) scanf("%d",&a)
#define sf64(a) scanf("%I64d",&a)
#define sf264(a,b) scanf("%I64d%I64d",&a,&b)
#define sf364(a,b,c) scanf("%I64d%I64d%I64d",&a,&b,&c)
#define sf464(a,b,c,d) scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&d)
#define sf2(a,b) scanf("%d%d",&a,&b)
#define sf3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define sf4(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)
#define sff(a) scanf("%f",&a)
#define sfs(a) scanf("%s",a)
#define sfs2(a,b) scanf("%s%s",a,b)
#define sfs3(a,b,c) scanf("%s%s%s",a,b,c)
#define sfd(a) scanf("%lf",&a)
#define sfd2(a,b) scanf("%lf%lf",&a,&b)
#define sfd3(a,b,c) scanf("%lf%lf%lf",&a,&b,&c)
#define sfd4(a,b,c,d) scanf("%lf%lf%lf%lf",&a,&b,&c,&d)
#define sfc(a) scanf("%c",&a)
#define ull unsigned long long
#define pp pair<int,int>
#define debug printf("***\n")
#define pi 3.1415927
#define mod 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
const double PI = acos(-1.0);
const double e = exp(1.0);
const int INF = 0x7fffffff;;
template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<class T> inline T Min(T a, T b) { return a < b ? a : b; }
template<class T> inline T Max(T a, T b) { return a > b ? a : b; }
bool cmpbig(int a, int b){ return a>b; }
bool cmpsmall(int a, int b){ return a<b; }
using namespace std;
bool cmp(pp a,pp b)
{
    if(a.first!=b.first)
        return a.first<b.first;
    return a.second<b.second;
}
ll a[500010];
ll lsum[500010];
int main()
{
    //freopen("data.in","r",stdin);
    //freopen("data.out" ,"w",stdout);
    ll n;
    while(cin>>n)
    {
        ll sum=0;
        rep(i,0,n)
        {
            sf64(a[i]);
            sum+=a[i];
            lsum[i]=sum;
        }
        if(sum%3!=0||n<3)
        {
            cout<<"0"<<endl;
            return 0;
        }
        ll cnt=0,pos=0;
        rep(i,0,n)
        {
            if(i!=n-1&&lsum[i]==sum*2/3)
                cnt+=pos;
            if(lsum[i]==sum/3)
                pos++;
        }
        cout<<cnt<<endl;
    }
    return 0;
}