H. Hard math problem

H. Hard math problem

Time Limit: 2000ms

Case Time Limit: 2000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

Submit Status PID: 29363

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    Xioumu was trapped by a mathematical puzzle recently.

 

    He wrote the number 0,1,2 on the blackboard, and then he does lots of operations on them, in each operation, he took out the larger two numbers of the three every time, added the two numbers and wrote down the sum on the blackboard, and then wiped any one of the original three numbers. He needed to use a minimum times of operations to get a maximum number x. Now he needs your help. 

Input

    The first line is an integer T indicates the number of test cases.

    Input contains several test cases. 

    For each test case, there is a number x (1<=x<=5*10^5) in a line.

Output

  For each case, output the case number first. Then output the minimum number of operations. If he can’t get x, just output -1.

Sample Input

3
2
6
13

Sample Output

Case 1: 0
Case 2: 4
Case 3: 4 

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

int ans;

int fff(int a,int b)

{

    ans++;

    if(a<1||b<2) return 0x3f3f3f3f;

    if(ans>=100) return 0x3f3f3f3f;

    if(a==1&&b==2) return ans;

    b=b-a;

    if(b<a)

        swap(a,b);

    fff(a,b);

}

int main()

{

    int T;

    cin>>T;

    int cas=1;

while(T--)

{

    int n;

    cin>>n;

    int cnt=0x3f3f3f3f;

    for(int i=1;i<=n/2;i++)

    {

        ans=0;

        int a=i;int b=n-i;

        if(a>b) swap(a,b);

        cnt=min(cnt,fff(a,b));

    }

    if(cnt>=0x3f3f3f3f&&n!=2) cnt=-1;

    if(n==2) cnt=0;

    printf("Case %d: %d\n",cas++,cnt);

}

    return 0;

}