http://codeforces.com/problemset/problem/466/C
题意:把数组分成三组连续的序列,每个序列总和相等;
思路:和为0特判。扫过去记录三分一sum[n]的个数,遇到三分二sum[n]的就加答案;
总结:细节没注意!
#include<bits/stdc++.h>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;
#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=5e5+9;
const int mod=1e9+7;
template <class T>
inline void sc(T &ret)
{
char c;
ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0'), c = getchar();
}
}
ll a[maxn];
ll sum[maxn];
map<ll ,ll>mp;
int main()
{
ll n;
cin>>n;
ll ans=0;
ll k=0;
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
for(int i=1;i<=n;i++)
{
sum[i]=sum[i-1]+a[i];
mp[sum[i]]++;
/*if(sum[i]%3==0&&sum[i]!=0)//2倍的可能在1倍后面,不行!!!
{
if(mp[sum[i]/3]>0&&mp[(sum[i]/3)*2]>0)
{
ans=ans+mp[sum[i]/3]*mp[(sum[i]/3)*2];
}
}*/
if(sum[i]==0)
{
k++;
}
}
ll num1=0;
if(sum[n]%3==0&&sum[n]!=0)
{
for(int i=1;i<=n;i++)
{
if(sum[i]==sum[n]/3) num1++;
if( num1 && (sum[i]==(sum[n]/3)*2) )
{
ans=ans+num1;
}
}
}
ll kk=0;
if(k>=3&&sum[n]==0)
{
if((k-1)%2==0)
kk=(k-1)/2,kk=kk*(k-2);
else
kk=(k-2)/2,kk=kk*(k-1);
}
cout<<ans+kk<<endl;
return 0;
}