标签：hduj c++
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<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.
注意输入就行了：
#include<cstring>
#include<iostream>
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<strlen(s1)-1;i++)
n*=10,n+=s1[i]-'0';
for(j=0;j<strlen(s2)-1;j++)
m*=10,m+=s2[j]-'0';
if(n==0) n=1;
if(m==0) m=1;
//printf("%d %d\n",n,m);
int sum,flag;
sum=flag=0;
for(i=0;i<=k;i+=n)
if((k-i)%m==0)
{
flag=1;
break;
}
if(flag) cout<<"Yes.\n\n";
else cout<<"No.\n\n";
}
return 0;
}
HUDJ 3270 The Diophantine Equation,布布扣,bubuko.com
HUDJ 3270 The Diophantine Equation
标签：hduj c++
原文地址：http://blog.csdn.net/hyccfy/article/details/38487815
