http://acm.hdu.edu.cn/showproblem.php?pid=3270
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<c<1000000. All other integers are positive and less than 100000. Note that, if a=1 then ‘ax’ will be represented as ‘x’ and same for by.
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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
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<strlen(s1)-1; i++)
{
if(s1[i]!='x')
{
a+=s1[i]-'0';
if(s1[i+1]!='x')
a*=10;
}
/* a*=10;
a+=s1[i]-'0';*/
}
for(j=0; j<strlen(s2)-1; j++)
{
if(s2[j]!='y')
{
b+=s2[j]-'0';
if(s2[j+1]!='y')
b*=10;
}
/* b*=10;
b+=s2[j]-'0';*/
}
if(s1[0]=='x')a=1;
if(s2[0]=='y')b=1;
if(qiujie(a,b,c)==1)
printf("Yes.\n\n");
else
printf("No.\n\n");
}
return 0;
}
杭电 3270 The Diophantine Equation