hdu-4180-exgcd

RealPhobia

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 938    Accepted Submission(s): 435

Problem Description

Bert is a programmer with a real fear of floating point arithmetic. Bert has quite successfully used rational numbers to write his programs but he does not like it when the denominator grows large. Your task is to help Bert by writing a program that decreases the denominator of a rational number, whilst introducing the smallest error possible. For a rational number A/B, where B > 2 and 0 < A < B, your program needs to identify a rational number C/D such that:
1. 0 < C < D < B, and
2. the error |A/B - C/D| is the minimum over all possible values of C and D, and
3. D is the smallest such positive integer.

 

 

Input

The input starts with an integer K (1 <= K <= 1000) that represents the number of cases on a line by itself. Each of the following K lines describes one of the cases and consists of a fraction formatted as two integers, A and B, separated by “/” such that:
1. B is a 32 bit integer strictly greater than 2, and
2. 0 < A < B

 

 

Output

For each case, the output consists of a fraction on a line by itself. The fraction should be formatted as two integers separated by “/”.

 

 

Sample Input

3 1/4 2/3 13/21

 

 

Sample Output

1/3 1/2 8/13

 

 

Source

The 2011 South Pacific Programming Contest

 

 　　  | A/B - C/D |= minn   <=>  | AD - BC| / BD =minn 

　　  如果AB可以约分的话直接约分就是答案。否则说明 gcd(A,B)=1, 我们有 A*D+B*C = gcd(A,B) = 1，原分子加了绝对值，有两种情况

D>0,C<0 或者是 D<0,C>0  ,解完之后对D分正负讨论一下那个使得分母更大就选那个，分子已经是1了。

　　因为D<B,所以记得%B，正负分别对应唯一的一个解。

　　

#include<iostream>
#include<cstdio>
using namespace std;
#define LL long long 
#define mp make_pair
#define pb push_back
#define inf 0x3f3f3f3f
void exgcd(LL a,LL b,LL &d,LL &x,LL &y){
    if(!b){d=a;x=1;y=0;}
    else{exgcd(b,a%b,d,y,x);y-=x*(a/b);}
}
int main(){
    LL a,b,d,x,y;
    int t;
    cin>>t;
    while(t--){
        scanf("%lld/%lld",&a,&b);
        exgcd(a,b,d,x,y);
        if(d!=1){
            printf("%lld/%lld\n",a/d,b/d);
        }
        else{
            LL d1,d2,c1,c2;
            d1=(x%b+b)%b,c1=-(1-a*d1)/b;
            d2=-(x%b-b)%b,c2=(1+a*d2)/b;
            if(d1>d2){
                printf("%lld/%lld\n",c1,d1);
            }
            else{
                printf("%lld/%lld\n",c2,d2); 
            }
        }
    }
    return 0;
}