time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You’ve got arraya[1],a[2],…,a[n], consisting ofnintegers. 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 indicesi,j(2≤i≤j≤n-1), that.
Input
The first line contains integern(1≤n≤5·105), showing how many numbers are in the array. The second line containsnintegersa[1],a[2], …,a[n](|a[i]|≤109)— the elements of arraya.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Sample test(s)
input
51 2 3 0 3
output
2
input
40 1 -1 0
output
1
input
24 1
output
0
默默的刷CF还是有些无聊，就写了好几个博客吐槽
话说这个题，还是很容易想到的，只是ai可正可负就有些蛋疼，不然直接二分很方便
在我看来，二分有两种写法，我对CF的机器很有信心，就用STL乱搞
反正都是nlogn，就是map的logn常数略大，哈哈哈，我有CF，不用怕
另外这个题的恰好分三段。。。简直就是一种模型，必然是前面弄一段，后面弄一段，，最后中间可以直接算……
我就是想到这个才想到正确的二分姿势……
#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<climits>#include<list>#include<iomanip>#include<stack>#include<set>using namespace std;typedef long long ll;map<ll,vector<int> >ps;ll p[int(5e5)+10];int main(){int n;cin>>n;ll sum=0;for(int i=1;i<=n;i++){cin>>p[i];sum+=p[i];ps[sum].push_back(i);}ll s=0,ans=0;for(int i=n;i>2;i–){s+=p[i];if(s*3==sum){map<ll,vector<int> >::iterator it;it=ps.find(s);int j=lower_bound((it->second).begin(),(it->second).end(),i)-(it->second).begin();ans+=j;if(j>0&&(it->second)[j-1]+1==(it->second)[j])ans–;}}cout<<ans;}
去旅行不在于记忆，而在于当时的那份心情。
