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  • HDU 2669 Romantic(裸的拓展欧几里得)

    Romantic

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 8536    Accepted Submission(s): 3638


    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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    lcy   |   We have carefully selected several similar problems for you:  2668 2674 2671 2670 2672 
     
     1 #include <iostream>  
     2 #include <cstdio>  
     3 #include <cstring>  
     4 #include <cmath>  
     5 #include <vector>  
     6 #include <string>  
     7 #include <queue>  
     8 #include <stack>  
     9 #include <algorithm>  
    10 
    11 #define INF 0x7fffffff  
    12 #define EPS 1e-12  
    13 #define MOD 1000000007  
    14 #define PI 3.141592653579798  
    15 #define N 100000  
    16 
    17 using namespace std;
    18 
    19 typedef long long ll;
    20 
    21 ll e_gcd(ll a, ll b, ll &x, ll &y)
    22 {
    23     ll d = a;
    24     if (b != 0)
    25     {
    26         d = e_gcd(b, a%b, y, x);
    27         y -= a / b * x;
    28     }
    29     else
    30     {
    31         x = 1; y = 0;
    32     }
    33     return d;
    34 }
    35 
    36 ll cal(ll a, ll b, ll c)
    37 {
    38     ll x, y;
    39     ll gcd = e_gcd(a, b, x, y);
    40     if (c%gcd != 0) return -1;
    41     x *= c / gcd;
    42     y *= c / gcd;
    43     if (b < 0) b = -b;
    44     ll ans = x % b;//最小的x,因为ax+by=c,尽量把ax分给b,多些y,少些x
    45     if (ans <= 0) ans += b;
    46     return ans;
    47 }
    48 
    49 int main()
    50 {
    51     ll a, b;
    52     while (cin>>a>>b)
    53     {
    54         ll ans = cal(a, b, 1);
    55         if (ans == -1) printf("sorry
    ");
    56         else printf("%I64d %I64d
    ", ans, (1 - ans * a) / b);
    57     }
    58     return 0;
    59 }
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  • 原文地址:https://www.cnblogs.com/caiyishuai/p/8454553.html
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