The Diophantine Equation hdu 3270

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The Diophantine Equation

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1130    Accepted Submission(s): 290

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.”

 

Author

Muhammed Hedayet

Source

HDOJ Monthly Contest – 2010.01.02 

题意：就是用扩展的欧几里得定理，如果有不懂得话就访问这个链接吧点击打开链接

既然知道姿势了就非常简单了可以把这个代码记下来，感觉一般求不定方程都一样：

#include <stdio.h>#include <string.h>char str[100];long long gcd(long long x,long long y){    return y==0?x:gcd(y,x%y);}long long gcd(long x,long y){    return y==0?x:gcd(y,x%y);}long long Ex_Euclid(long long a,long long b,long long &x,long long &y){    long long ans,t;    if (b==0)    {        x=1;        y=0;        ans=0;        return ans;    }    ans=Ex_Euclid(b,a%b,x,y);    t=x;    x=y;    y=t-(a/b)*y;    return ans;}int main(){    int i,j,n;    long long A,B,C,D,x,y,k,t;    while(scanf("%s",str)!=EOF)    {        A=0;        for (i=0;i<strlen(str)-1;i++)        {            A=A*10+str[i]-'0';        }        scanf("%s",str);        scanf("%s",str);        B=0;        for (i=0;i<strlen(str)-1;i++)        {            B=B*10+str[i]-'0';        }        scanf("%s",str);        scanf("%s",str);        C=0;        for (i=0;i<strlen(str);i++)        {            C=C*10+str[i]-'0';        }        if (A==0) A=1;        if (B==0) B=1;        D=gcd(A,B);        if (C%D!=0)        {            printf("No.\n\n");            continue;        }        n=Ex_Euclid(A,B,x,y);        x=x*C/D;        t=B/D;        x=(x%t+t)%t;        k=(C-A*x)/B;        if (k>=0)        {            printf("Yes.\n\n");            continue;        }        y=y*C/D;        t=A/D;        y=(y%t+t)%t;        k=(C-B*y)/A;        if (k>=0)        {            printf("Yes.\n\n");            continue;        }        printf("No.\n\n");    }    return 0;}

这个题除了用扩展的欧几里得还可以用直接暴搜，感觉是不是很暴力：

#include<cstdio>#include<cstring>#include<cstdlib>char arr[30],temp[10];int a,b,c;int main(){    int i,j,len,x,y,cnt;    while(scanf("%s",temp)!=EOF)    {        getchar();getchar();getchar();        a=atoi(temp);        if(a==0)            a=1;            scanf("%s",temp);            getchar();getchar();getchar();            b=atoi(temp);            if(b==0)                b=1;            scanf("%d",&c);            getchar();            if(c>=0)            {                if(b==0&&c%a==0||a==0&&c%b==0)                   {                        printf("Yes.\n\n");                        break;                   }                   for(x=0;(cnt=c-a*x)>=0;x++)                   {                       if(cnt%b==0)                        {                          printf("Yes.\n\n");                            break;                        }                   }                   if(cnt<0)                     printf("No.\n\n");            }            else                 printf("No.\n\n");    }    return 0;}

稍稍的解释一下公式吧：ax+by=c ;

y=(c-ax)/b;记住分子要大于0，很简单吧。