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！！！ |
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#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=0; for(y=0;y<=n/2;y++){ sum+=(n-2*y)/2+1; } cout<<sum<<'\n'; } }
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