codeforces round #266 C. Number of Ways

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本文介绍了一种算法，用于计算一个整数数组能够被划分为三个连续子数组部分的方法数量，这三个部分的元素总和相等。讨论了输入输出格式、时间及内存限制，并通过样例展示了算法的应用。

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.

Sample test(s)

input
5
1 2 3 0 3

output
2

input
4
0 1 -1 0

output
1

input
2
4 1

output
0

统计有多少组i,j，使得sum[1,i].sum[i,j],sum[j+1,n]相等

考虑的东西太少了，fst了。。

不说了。。刚开始少考虑了2*sum/3 可能在 sum/3后面的。。

/***********************************************\
 |Author: YMC
 |Created Time: 2014/9/13 0:15:56
 |File Name: c.cpp
 |Description: 
\***********************************************/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
#define mset(l,n) memset(l,n,sizeof(l))
#define rep(i,n) for(int i=0;i<n;++i)
#define maxx(a) memset(a, 0x3f, sizeof(a))
#define zero(a) memset(a, 0, sizeof(a))
#define srep(i,n) for(int i = 1;i <= n;i ++)
#define MP make_pair
const int inf=0x3f3f3f3f ;
const double eps=1e-8 ;
const double pi=acos (-1.0);
typedef long long ll;

using namespace std;
ll da[500005];
int n;
ll sum;
int main() {
	//freopen("input.txt","r",stdin); 
    scanf("%d",&n);
    sum = 0;
    rep(i,n){
        scanf("%I64d",&da[i]);
        sum += da[i];
    }
    if(sum == 0){
        ll ss = 0;
        ll ss1 = 0;
        for(int i=0;i<n;++i){
            ss += da[i];
            if(ss == 0) ss1 ++;
        }
        if(ss1 == 1 || ss1 == 2) {
            cout<<0<<endl;
            return 0;
        } else {
            cout<<(1+ss1-2)*(ss1-2)/2<<endl;
            return 0;
        }
    } 
    ll tt = sum / 3;
    if(tt * 3 != sum){
        puts("0");
        return 0;
    }
    ll t1 = tt;ll t2 = tt*2;
    ll tt1 = 0,tt2 = 0;
    ll su = 0;
    ll cc = 0;
    ll ans = 0;
    rep(i,n){
        su += da[i];
        if(su == t2) ans += cc;
        if(su == t1) cc ++;
        //if(su == t2) tt2 ++; 
    }
    cout<<ans<<endl;
	return 0;
}