[ACM_数学] Counting Solutions to an Integral Equation (x+2y+2z=n 组合种类)

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27938#problem/E

题目大意:Given, n, count the number of solutions to the equation x+2y+2z=n, where x,y,z,n are non negative integers.
解题思路:只对一个枚举由此推算出另外两个的种类,千万不要都枚举!!!
#include<iostream>
#include<fstream>
#include<string>
#include<string.h>
using namespace std;
int main(){
for(long long int n;cin>>n;){
long long int y,sum=;
for(y=;y<=n/;y++){
sum+=(n-*y)/+;
}
cout<<sum<<'\n';
}
}
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