hdu2601 an easy problem (mathematics)

Topic Meaning:

http://acm.hdu.edu.cn/showproblem.php?pid=2601

Give a number n, ask N=i*j+i+j altogether how many kinds of scheme.

Topic Analysis:

The direct violent simulation of this problem, just need to convert the n+1= (i+1) * (j+1) to calculate.

AC Code:

/** *hdu2601 an easy problem * meaning: To find the type of N=I*J+I+J * Analysis: Convert to N+1= (i+1) * (j+1), solve on the line * * #include <iostream> #include < cstdio> #include <map> #include <cstring> #include <string> #include <algorithm> #include <queue> #include <vector> #include <stack> #include <cstdlib> #include <cctype> #include <cstring> #include <cmath>using namespace Std;int main () {    int t;long long n;    cin>>t;    while (t--) {        cin>>n;        n=n+1;        int k=0;        For (long long i=2;i*i<=n;i++) {            if (n%i==0) k++;        }        cout<<k<<endl;    }    return 0;}

hdu2601 an easy problem (mathematics)