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本文探讨了一种算法问题，即如何将一个整数数组分割成三个连续的部分，使得每个部分的元素之和相等。文章提供了两种解决方案，一种是通过遍历所有可能的分割点并检查条件，但这种方法效率较低；另一种是优化后的算法，利用前缀和的概念来减少计算复杂度。

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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

Copy

5
1 2 3 0 3

output

Copy

input

Copy

4
0 1 -1 0

output

Copy

input

Copy

2
4 1

output

Copy

题意：一段数组，分三块“和”相同的块，问有多少种分法？

思路：用前缀和

#include<cstdio>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cmath>
#include<vector>
#include<cstring>
#include<string>
#include<iostream>
#include<iomanip>
#define mset(a,b)   memset(a,b,sizeof(a))
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
//const ll mod=1e9+7;
const int N=1000000;
const int inf=0x3f3f3f3f;
//priority_queue<int,vector<int>,greater<int> >q;
ll a[500005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        ll sum=0;
        a[0]=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
            a[i]+=a[i-1];
        }
//        for(int i=0;i<n;i++)
//        printf("%lld ",a[i]);printf("\n");
        if(a[n-1]%3)
            printf("0\n");
        else
        {
            for(int i=0;i<n-1;i++)
            {
                if(a[i]*3==a[n-1])
                {
                    for(int j=i+1;j<n-1;j++)
                    {

                        if(a[j]*3==a[n-1]*2)
                        {
                            sum++;
                        }

                    }
                }

            }
            printf("%lld\n",sum);
        }
    }
    return 0;
}

上面爆搜…………超时…………我也很无奈啊，然后是下面

#include<cstdio>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cmath>
#include<vector>
#include<cstring>
#include<string>
#include<iostream>
#include<iomanip>
#define mset(a,b)   memset(a,b,sizeof(a))
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
//const ll mod=1e9+7;
const int N=1000000;
const int inf=0x3f3f3f3f;
//priority_queue<int,vector<int>,greater<int> >q;
ll a[500005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        ll sum=0;
        ll ans=0;
        a[0]=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
            a[i]+=a[i-1];
        }
//        for(int i=0;i<n;i++)
//        printf("%lld ",a[i]);printf("\n");
        if(a[n-1]%3)
            printf("0\n");
        else
        {
            for(int i=0;i<n-1;i++)
            {
                if(a[i]*3==a[n-1]*2)
                    sum+=ans;//第二块和第三块
                if(a[i]*3==a[n-1])
                    ans++;//第一块和第三块
            }
            printf("%lld\n",sum);
        }
    }
    return 0;
}