杭电 3270 The Diophantine Equation

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 = 1015x + 35y = 67x + 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;}