joj1012

1012: Diophantus was a Greek, not a geek.

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	4695	1754	Standard

An equation of the form ax + by = c is a Diophantine equation, where a, b, and c are positive integers.

Such an equation has no integer solution if the greatest common divisor of a and b does not divide c. If it does divide c, then a solution exists.

The greatest common divisor (gcd) of two positive integers, is the largest integer that divides both a and b. For example, the gcd of 2 and 6 is 2. The gcd of 9 and 12 is 3. The gcd of 15 and 7 is 1.

Given a diophantine equation, determine if it has any solutions.

Input

A positive integer, denoting the number of equations.

A diophantine equation. Each equation will be of the form: <num>x+<num>y=<num>, where <num> denotes a positive integer(<=65536). Each equation will be on a separate line and will contain no blanks.

Output

The phrase “<equation> has a solution.” if a solution exists, or the phrase “<equation> has no solution.” if one does not exist, where <equation> is the equation in question.

Sample Input

2
2x+5y=11
60x+18y=97

Sample Output

2x+5y=11 has a solution.
60x+18y=97 has no solution.

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This problem is used for contest: 53 

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Problem Set with Online Judge System Version 3.12
Jilin University

Developed by skywind, SIYEE

 

 

 

 

 

#include<cstdio>
#include<ctype.h>
/*int gcd(int m,int n)
{
    if(m%n==0)return n;
    else gcd(n,m%n);
}*/
int gcd(int m,int n)
{
    int s=1;
    while(s!=0)
    {
        s=m%n;
        m=n;
        n=s;
    }
    return m;
}
int integer(char *p,int n)
{
    int sum=0;
    int u=1;
    for(int i=n-1;i>=0;i--)
    {
        sum+=(p[i]-'0')*u;
        u=u*10;
    }
    return sum;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    char a[100];
    char b[10];
    int c[3];
    scanf("%d",&n);
    getchar();
    for(int i=0;i<n;i++)
    {
        int cc=0;
        gets(a);
        int count=0;
        for(int j=0;a[j]!='\0';j++)
        {
            if(isdigit(a[j]))
            {
                b[count]=a[j];
                count++;
            }
            if(!isdigit(a[j])&&count!=0)
            {
                c[cc]=integer(b,count);
                ++cc;
                count=0;
            }
        }
        c[2]=integer(b,count);
        if((c[2]%gcd(c[0],c[1]))==0)
        {
            printf("%s has a solution.\n",a);
        }
        else
        {
            printf("%s has no solution.\n",a);
        }
    }
    return 0;
}