2200: Fracions

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本文介绍了一个算法问题：将分数a/b(a<b)以1/b1+1/b2+...+1/bn的形式表达，并使b1到bn之和最小。文章提供了输入输出样例及解析方法。

2200: Fracions

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	292	107	Standard

A fracion a/b(a < b) can be expressed in the way 1/b1 + 1/b2 + ... + 1/bn. Now can you achieve it and make the sum of b1 to bn minimum?

Input and Output

For each case there are tow positive intergers a and b( 0 < a < b <= 100), output the minimum sum.

Sample Input

2 3
3 4
4 15

Sample Output

6
6
16

Hint: 2/3 = 1/3 + 1/3, 4/15 = 1/6 + 1/10

This problem is used for contest: 29

#include<stdio.h>
int gcd(int x,int y) { if(y==0) return x;return gcd(y,x%y);}
int n,m,min;
void dfs(int id,int nown,int nowm,int nowsum)
{
 // printf("%d...%d....%d....%d/n",id,nown,nowm,nowsum);
 if(nowsum>min) return ;
 if(id>min) return ;
 if(nown*m>nowm*n) return ;
 if(nown*m==nowm*n) {if(nowsum<min) min=nowsum;return ;}
 dfs(id+1,nown*id+nowm,id*nowm,nowsum+id);
 dfs(id,nown*id+nowm,id*nowm,nowsum+id);
 dfs(id+1,nown,nowm,nowsum);
}
int main()
{
 //freopen("out.txt","w",stdout);
 while(scanf("%d%d",&n,&m)==2)
 {
 int t=gcd(n,m);n/=t,m/=t;
 min=n*m;
 dfs(2,0,1,0);
 printf("%d/n",min);
 }
 return 0;
}