题目链接：http://codeforces.com/problemset/problem/466/C
题目：
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
51 2 3 0 3
output
2
input
Copy
40 1 -1 0
output
1
input
Copy
24 1
output
0
题解：前缀和暴力，找下规律。（好像还能用DP解，明天起来再看看！）
1 #include <map>
2 #include <cstdio>
3 #include <iostream>
4 #include <algorithm>
5 using namespace std;
6
7 typedef long long LL;
8 const int N=1e6;
9 LL f[N],c1[N],c2[N];
10
11 int main(){
12 LL n,ans=0,cnt=0;
13 scanf("%lld",&n);
14 for(int i=1;i<=n;i++){
15 scanf("%lld",&f[i]);
16 f[i]+=f[i-1];
17 if(!f[i]) cnt++;
18 }
19 if(f[n]%3!=0) {printf("0\n");return 0;}
20 if(f[n]==0){
21 printf("%lld\n",(cnt-1)*(cnt-2)/2);
22 return 0;
23 }
24
25 for(int i=1;i<=n;i++){
26 c1[i]=c1[i-1];c2[i]=c2[i-1];
27 if(f[i]==(f[n]/3)) c1[i]++;
28 if(f[i]==(f[n]/3*2)) c2[i]++;
29 }
30 for(int i=2;i<n;i++){
31 if(f[i]==(f[n]/3*2)) ans+=c1[i];
32 }
33 printf("%lld\n",ans);
34 return 0;
35 }
原文地址：https://www.cnblogs.com/Leonard-/p/8503515.html