Xtuoj ABK (finding out the number of major conventions on A and B)

Accepted:21		submit:171
Time limit:1000 MS		Memory limit:65536 KB

Title Description

ABK is a more simple topic than a+b, give two integers, a, B, to find out the number of the k conventions of A and C.

Input

The first line is an integer N (n≤10000) that represents the number of samples. For each subsequent line, a sample of 3 integers a,b,k (1≤a,b≤109, 1≤k≤10)

Output

Each line outputs an integer d, which indicates that the number of major K conventions of A/b is output-1 without the number of conventions of the K-large.

Ideas: First find out A and B gcd,a and B of the number of conventions must also be greatest common divisor, so the number of K conventions can be from 1 to GCD enumeration K, the GCD divided by the K small approximate is the K large approximate, but 10^9 the amount of data will be t. So can be from 1 to the root of the GCD enumeration, assuming that the name of N, if the n>=k will be found, if n<k, the remaining k-n must be in (gcd-square root gcd) within the range, can not be enumerated, otherwise t, to use good approximate nature, So it can still be obtained by gcd the square root to 1 Inverse enumeration k-n gcd, the complexity o (square root n).

The code is as follows:

 #include < iostream> #include <cstdio> #include <cstring> #include <string> #include <map> #include < algorithm> #include <set> #include <cmath> #include <vector>using namespace Std;int a,b,k;int gcd (    int A,int b) {if (b==0) return A; else return gcd (b,a%b);}    void Solve () {int mx=gcd (A, b), cnt=0;        for (int i=1;i*i<=mx;i++)//Forward enumeration if (mx%i==0) {cnt++;     if (cnt==k) cout<<mx/i<<endl;        } for (int i=floor (sqrt (MX) +0.5); i>=1;i--)//Reverse enumeration//floor in order to avoid error if (mx%i==0) {cnt++;        if (I*I==MX) cnt--;    Small details, if gcd's arithmetic square root is exactly an integer, the first I of the inverse enumeration and the last I of the forward enumeration repeat if (CNT==K) cout<<i<<endl; } if (cnt<k) cout<< "-1" <<ENDL;}   int main () {int _;   cin>>_;       while (_--) {cin>>a>>b>>k;   Solve (); } return 0;} 

Xtuoj ABK (finding out the number of major conventions on A and B)