• HDU 2669 Romantic


    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 6409    Accepted Submission(s): 2667


    Problem Description
    The Sky is Sprite.
    The Birds is Fly in the Sky.
    The Wind is Wonderful.
    Blew Throw the Trees
    Trees are Shaking, Leaves are Falling.
    Lovers Walk passing, and so are You.
    ................................Write in English class by yifenfei



    Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
    Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead.
     
    Input
    The input contains multiple test cases.
    Each case two nonnegative integer a,b (0<a, b<=2^31)
     
    Output
    output nonnegative integer X and integer Y, if there are more answers than the X smaller one will be choosed. If no answer put "sorry" instead.
     
    Sample Input
    77 51 10 44 34 79
     
    Sample Output
    2 -3 sorry 7 -3
     
    Author
    yifenfei
     
    Source
     
     
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    裸的扩展欧几里得
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<algorithm>
     6 #define lli long long int 
     7 using namespace std;
     8 void read(lli &n)
     9 {
    10     char c='+';lli x=0;bool flag=0;
    11     while(c<'0'||c>'9')
    12     {c=getchar();if(c=='-')flag=1;}
    13     while(c>='0'&&c<='9')
    14     {x=x*10+(c-48);c=getchar();}
    15     flag==1?n=-x:n=x;
    16 }
    17 lli x,y;
    18 lli exgcd(lli a,lli b,lli & x,lli & y)
    19 {
    20     if(b==0)
    21     {
    22         x=1;y=0;
    23         return a;
    24     }
    25     lli r=exgcd(b,a%b,y,x);
    26     y-=a/b*x;
    27     return r;
    28 }
    29 int main()
    30 {
    31     lli a,b;
    32     while(scanf("%lld%lld",&a,&b)==2)
    33     {
    34         lli d=exgcd(a,b,x,y);
    35         if(1%d!=0)
    36             printf("sorry
    ");
    37         else if(d==-1)
    38             printf("sorry
    ");
    39           else 
    40         {
    41             x*=1/d;
    42               b=fabs(b);
    43               b=b/d;
    44             lli ans=x%b;
    45             if(ans<=0)
    46                 ans+=b;
    47             printf("%lld %lld
    ",ans,(1-a*ans)/b);        
    48         }
    49     }
    50     
    51     return 0;
    52 }
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/7100082.html
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