hdu-2669(扩展欧几里得）

Problem Description

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 5110 4434 79

Sample Output

2 -3sorry7 -3

题目题意十分明显，就是让你求x*a+y*b=1,这个不定方程，x的解有一个条件就是，最小的正整数。

题目分析:扩展欧几里得算法可以解决这样的不定方程x*a+y*b=c，对于这样的一个方程.

我们依据的是最初的方程a*x+b*y=gcd(a,b) ，当c|gcd(a,b)时方程有解。

bool linear_equation(int a,int b,int c,int &x,int &y){    int d=exgcd(a,b,x,y);    if(c%d)//未满足要求        return false;    int k=c/d;    x*=k; y*=k;    //求得的只是其中一组解    return true;}ax+by=c其他的解满足:X=x1+b/gcd(a,b)*t;//方程所有的解Y=y1-a/gcd(a,b)*t;其中t为任意整数

代码如下:

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int gcd(int a,int b){    if (b==0) return a;    else gcd(b,a%b);}int exgcd(int a,int b,int &x,int &y)//解决ax+by=gcd(a,b){    if (b==0) {        x=1;        y=0;        return a;    }    int r=exgcd(b,a%b,x,y);    int t=x;    x=y;    y=t-a/b*y;    return r;}bool linear_equation(int a,int b,int c,int &x,int &y){    int d=exgcd(a,b,x,y);    if (c%d)        return false;    int k=c/d;    x*=k,y*=k;    int t=0;    int temp=b/d;    if (x<0) {//求最小的整数x        while (x<0) {            t++;            x=x+temp*t;        }        printf("%d %d\n",x,y-a/d*t);    }    else {        while (x>0) {            t--;            x=x+temp*t;        }        printf("%d %d\n",x+temp,y-a/d*(t+1));    }    return true;}int main(){    int a,b;    while (scanf("%d%d",&a,&b)!=EOF) {        int x,y;        if (!linear_equation(a,b,1,x,y))            printf("sorry\n");    }    return 0;}