hdu3270 The Diophantine Equation

The Diophantine Equation

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1116    Accepted Submission(s): 287

Problem Description

We will consider a linear Diaphonic equation here and you are to find out whether the equation is solvable in non-negative integers or not.
Easy, is not it?

 

Input

There will be multiple cases. Each case will consist of a single line expressing the Diophantine equation we want to solve. The equation will be in the form ax + by = c. Here a and b are two positive integers expressing the co-efficient of two variables x and y.
There are spaces between:
1. “ax” and ‘+’
2. ‘+’ and “by”
3. “by” and ‘=’
4. ‘=’ and “c”

c is another integer that express the result of ax + by. -1000000

 

Output

You should output a single line containing either the word “Yes.” or “No.” for each input cases resulting either the equation is solvable in non-negative integers or not. An equation is solvable in non-negative integers if for non-negative integer value of x and y the equation is true. There should be a blank line after each test cases. Please have a look at the sample input-output for further clarification.

 

Sample Input

2x + 3y = 10 15x + 35y = 67 x + y = 0

 

Sample Output

Yes. No. Yes. HINT: The first equation is true for x = 2, y = 2. So, we get, 2*2 + 3*2=10. Therefore, the output should be “Yes.”

 一开始没有看到非负解，写了个错误的代码，只判断是否有解。。。。

代码一：

#include
#include
#include
int exgcd(int a,int b,int &x,int &y)
{
    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;
    return true;
}
int main()
{
    int c,x,y,N,M;
    char a[10],b[10];
    while(scanf("%s + %s = %d",a,b,&c)!=EOF)
    {
        N=atoi(a);
        if(N==0)N=1;
        M=atoi(b);
        if(M==0)M=1;
        if(linear_equation(N,M,c,x,y))
            printf("Yes.\n\n");
        else
            printf("No.\n\n");
    }
    return 0;
}

代码二：（正解！）

#include
#include
long long gcd(long long x,long long y)
{
    return y==0?x:gcd(y,x%y);
}
long long exgcd(long long a,long long b,long long &x,long long &y)
{
    long long r,t;
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    r=exgcd(b,a%b,x,y);
    t=x;
    x=y;
    y=t-(a/b)*y;
    return r;
}
int main()
{
    long long c,x,y,N,M,t,d,n,k;
    char a[100],b[100];
    while(scanf("%s + %s = %lld",a,b,&c)!=EOF)
    {
        N=atoi(a);
        if(N==0)N=1;
        M=atoi(b);
        if(M==0)M=1;
        d=gcd(N,M);
        if(c%d!=0)
        {
            printf("No.\n\n");
            continue;
        }
        n=exgcd(N,M,x,y);
        x=x*c/d;
        t=M/d;
        x=(x%t+t)%t;
        k=(c-N*x)/M;
        if(k>=0)
        {
            printf("Yes.\n\n");
            continue;
        }
        y=y*c/d;
        t=N/d;
        y=(y%t+t)%t;
        k=(c-M*y)/N;
        if (k>=0)
        {
            printf("Yes.\n\n");
            continue;
        }
        printf("No.\n\n");
    }
    return 0;
}