H.I.T_2815---扩展欧几里德-优快云博客

本文探讨了如何通过加减操作使两个整数得到1的最少步骤问题，并提供了使用扩展欧几里德算法的实现代码。针对不同输入，算法能够找到达到目标所需的最短路径。

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Min Chain

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	Source : 2009 China North East Local Contest
	Time limit : 2 sec		Memory limit : 64 M

Submitted : 435, Accepted : 117

Problem Description

Raven likes games of numbers. Today he meets two numbers and thinks whether he could get a result of 1 by doing at least one operation (addition or subtraction). However, he is tired of calculation; he also wants to know the minimum steps of operation that he could get 1.

Input Details

The first line of the input contains an integer T, which indicates the number of test cases.

In the following T rows, there are two positive integers a, b ( 0<=a, b<=10^9) in each row.

Output Details

For each case, output the least number of steps.

If you cannot get 1, just output -1.

Sample Input

Sample Output

1
10
-1

Hint

Sample 1: 3 - 2 = 1， One subtraction will be needed.
Sample 2: 16-9+16-9+16-9-9-9+16-9-9=1，It requires 10 additions and subtractions.
Sample 3: You cannot get 1.
此题还是看学长博客才发现自己漏掉好多情况，而且对扩展欧几里德算法没有真正搞明白，自己太没有钻研精神啊--。
#include <iostream>
#include <math.h>
using namespace std;
void ex(long long a,long long b,long long &x,long long &y)
{
    if(b==0)
    {
        x=1;
        y=0;
       return;
    }
    ex(b,a%b,x,y);
    long long tmp=x;
    x=y;
    y=tmp-(a/b)*y;
}
long long gcd(long long a,long long b)
{
    if(b==0)
    return a;
    return gcd(b,a%b);
}
int main()
{
    long long a,b,x,y;
    long long ans;
    int t;
    cin>>t;
    while(t--)
    {  cin>>a>>b;
    if(a<b)
    swap(a,b);
       if(gcd(a,b)!=1)
       {cout<<"-1"<<endl;
       continue;
       }
          if(b==1&&a==2)
            {
              cout<<"1"<<endl;
              continue;
            }
            if(b==1)
            {
               cout<<"2"<<endl;
               continue;
            }
            if(b==0&&a==1)
            {
                cout<<"1"<<endl;
                continue;
            }
           ex(a,b,x,y);
           ans=fabs(x)+fabs(y);
           if(x<0)
           {
               long long tmp=fabs(x+b)+fabs(y-a);
               if(tmp<ans)
               ans=tmp;
           }
           else
           {
               long long tmp=fabs(x-b)+fabs(y+a);
               if(tmp<ans)
               ans=tmp;
           }
           cout<<ans-1<<endl;

    }
    //cout << "Hello world!" << endl;
    return 0;
}