The Diophantine Equation

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1146 Accepted Submission(s): 297

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.”

一道很简单的题目，稍稍细心一些就能很快的AC，（我愚钝呀，弄了好久，后来想想觉得A出来之前大脑是短路状态，，这是为什么啊）需要注意的是输入，处理的话也不是很难。 AC代码：

#include
   
    
#include
    
      #include
     
       #include
      
        using namespace std; int qiujie(int a,int b,int c) { int x; for(x=0; x<=c/a; x++) //枚举x可取的所有可能 { if((c-a*x)%b==0) //判断是否存在正整数y { return 1; } } return 0; } int main() { int a,b,c,i,j; char s1[16],s2[16],ch,cj; while(cin>>s1>>ch>>s2>>cj>>c) { a=b=0; for(i=0; i


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