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杭电 3270 The Diophantine Equation

2024-07-17 01:28:58   219人阅读

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，，这是为什么啊快哭了）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;
}


 


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