codeforces 466C Number of Ways_codeforces c. number of ways-优快云博客

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/77898669

本文介绍了一种将数组等分为三个连续子数组的方法，每个子数组的元素总和相等。通过两道示例题目详细解释了算法实现，并提供了两种不同的代码实现方式。

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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 $\text{[math]}$ .

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

题意：将这个数组分成三段，每段的和相等，求有几种分法。

我自己没想出来，这个是我比较喜欢的一种

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
const int ma=5*1e5+10;
typedef long long ll;
using namespace std;
ll a[ma];
ll s=0,n;
vector<int>vec;
ll solve()
{
    if(s%3)
        return 0;
    s/=3;
    ll p=0;
    for(int i=0; i<n; i++)
    {
        p+=a[i];
        if(p==s)
            vec.push_back(i);
    }
    p=0;
    ll ans=0;
    for(int i=n-1;i>=0;i--)
    {
        p+=a[i];
        if(p==s)
            ans+=lower_bound(vec.begin(),vec.end(),i-1)-vec.begin();
    }
    return ans;
}
int main()
{
    cin>>n;
    for(int i=0; i<n; i++)
    {
        cin>>a[i];
        s+=a[i];
    }
    cout<<solve()<<endl;
    return 0;
}

这个也能过，但是不明白all为0时，那个几种为什么是 (cnt-1)*(cnt-2)/2这个，如果你知道，欢迎评论，感激不尽。

#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int ma=500005;
ll a[ma],sum[ma];
int main()
{
    int n;
    cin>>n;
    sum[0]=0;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        sum[i]=sum[i-1]+a[i];
    }
    if(n==1)
        cout<<0<<endl;
    else
    {
        ll all=sum[n];
        if(all%3!=0)
            cout<<0<<endl;
        else if(all)
        {
            ll s=0;
            ll w1=all/3;
            ll w2=all/3*2;
            int cnt=0;
            for(int i=1;i<=n;i++)
            {
                if(sum[i]==w2)
                    s+=cnt;
                if(sum[i]==w1)
                    cnt++;
            }
            cout<<s<<endl;
        }
        else
        {
            ll cnt=0;
            for(int i=1;i<=n;i++)
            {
                if(sum[i]==0)
                    cnt++;
            }
            cout<<(cnt-1)*(cnt-2)/2<<endl;
        }
    }
    return 0;
}