题目链接
题意:
给出你一个序列让你三等分,求一共多少种分法.
思路:
首先判断和是否能整除3,然后维护一个前缀和,之后三等分,两个断点就是1倍的sum/3,和2倍的sum/3,找出这两个断点.
#include <set>
#include <map>
#include <queue>
#include <string>
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
typedef pair<ll, ll> pii;
#define mem(a,x) memset(a,x,sizeof(a))
#define debug(x) cout << #x << ": " << x << endl;
#define rep(i,n) for(int i=0;i<(n);++i)
#define repi(i,a,b) for(int i=int(a);i<=(b);++i)
#define repr(i,b,a) for(int i=int(b);i>=(a);--i)
const int maxn = 5e5 + 1010;
#define inf 0x3f3f3f3f
#define sf scanf
#define pf printf
const int mod = 998244353;
const int MOD = 10007;
inline int read()
{
int x = 0;
bool t = false;
char ch = getchar();
while((ch < '0' || ch > '9') && ch != '-')ch = getchar();
if(ch == '-')t = true, ch = getchar();
while(ch <= '9' && ch >= '0')x = x * 10 + ch - 48, ch = getchar();
return t ? -x : x;
}
/*
vector<ll> m1;
vector<ll> m2;
priority_queue<ll , vector<ll> , greater<ll> > mn;//上 小根堆 小到大
priority_queue<ll , vector<ll> , less<ll> > mx;//下 大根堆 大到小
*/
map<ll, ll>mp;
map<ll, ll>mpp;
multiset<ll> s, t;
vector<ll> ans;
ll n, m, u, d, r, l;
ll a[maxn], b[maxn];
#define read read()
int main()
{
cin >> n;
for(int i = 1; i <= n; i++) scanf("%lld", &a[i]);
for(int i = 1; i <= n; i++)
{
b[i] = a[i] + b[i - 1];
}
if(b[n] % 3) puts("0");
else
{
ll ans = b[n] / 3;
ll num = 0, num1 = 0, num2 = 0;
for(int i = 1; i < n; i++)
{
if(ans * 2 == b[i])
{
num1 = num1 + num;
}
if(ans == b[i])
{
num++;
}
}
cout << num1 << endl;
}
return 0;
}