zoukankan      html  css  js  c++  java
  • HDU 4180 扩展欧几里得

    RealPhobia

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 376    Accepted Submission(s): 151


    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
     
    Recommend
    lcy   |   We have carefully selected several similar problems for you:  4186 4181 4182 4183 4179
     
     
     
    #include<stdio.h>
    #include<string.h>
    
    long long gcd1(long long a,long long b,long long &x,long long &y)
    {
        if(b == 0)
        {
            x = 1;
            y = 0;
            return a;
        }
        long long d = gcd1(b,a%b,x,y);
        long long t = x;
        x = y;
        y = t - a/b*y;
        return d;
    }
    
    
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            long long a,b;
            scanf("%lld/%lld",&a,&b);
            long long x = 0,y = 0;
            long long p = gcd1(a,b,x,y);
            //printf("%lld,%lld
    ",x,y);
            //printf("---%lld
    ",p);
            //printf("==%lld %lld
    ",a,b);
            if(p != 1)
            {
                printf("%lld/%lld
    ",a/p,b/p);
                continue;
            }
            if(a == 1)
            {
                printf("1/%lld
    ",b-1);
                continue;
            }
            long long x1 = 0,y1 = 0;
            if(x > 0)
            {
                x1 = (a + y)%a;
                y1 = (b - x)%b;
            }
            else
            {
                x1 = (a - y)%a;
                y1 = (b + x)%b;
            }
            //printf("%lld %lld %lld %lld
    ",x1,y1);
            printf("%lld/%lld
    ",x1,y1);
        }
        return 0;
    }
  • 相关阅读:
    CF Gym 101955G Best ACMer Solves the Hardest Problem 平面加点,删点,点加权(暴力)
    CF Gym 101955C Insertion Sort
    狩猎大赛(社团周赛)
    CF 1215 D Ticket Game (博弈)
    CF1215 C Swap Letters(思维题)
    CF 1215 B The Number of Products(思维题)
    HDU 6740 MUV LUV EXTRA(求循环节)
    BZOJ 1491 [NOI2007]社交网络(Floyd变形)
    BZOJ 3039 玉蟾宫
    【luogu4124】【bzoj4521】 [CQOI2016]手机号码 [数位dp]
  • 原文地址:https://www.cnblogs.com/13224ACMer/p/4729081.html
Copyright © 2011-2022 走看看