HUDJ 3270 The Diophantine Equation

The Diophantine Equation

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1133 Accepted Submission(s): 292


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.

注意輸入就行了：

#include
#include
using namespace std;

int main()
{
    char s1[20],s2[20];char q,w;
    int k;
    while(cin>>s1>>q>>s2>>w>>k)
    {
        int n,m,i,j;
        n=m=0;
        for(i=0;i