HDU 3270 The Diophantine Equation

The Diophantine Equation

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1285    Accepted Submission(s): 369

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

 

Recommend

chenheng

题意：判断ax+by=c是否存在x y位非负数解

可以用扩展欧几里得求出一个最小x，根据y=（c-ax）/b求得，再判断x y是否同时位非负数就行，同理计算一个最小y

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<stack>#include<queue>#include<deque>#include<set>#include<map>#include<cmath>#include<vector>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int, int> PII;#define pi acos(-1.0)#define eps 1e-10#define pf printf#define sf scanf#define lson rt<<1,l,m#define rson rt<<1|1,m+1,r#define e tree[rt]#define _s second#define _f first#define all(x) (x).begin,(x).end#define mem(i,a) memset(i,a,sizeof i)#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)#define mi ((l+r)>>1)#define sqr(x) ((x)*(x))const int inf=0x3f3f3f3f;char g[100];ll A,B,C,D,x,y;ll e_gcd(ll a,ll b,ll& x,ll& y){    if(!b)    {        x=1,y=0;        return a;    }    ll ans=e_gcd(b,a%b,x,y),tmp=x;    x=y,y=tmp-a/b*y;    return ans;}int main(){    while(~sf("%s",g))    {        int tag=0;        A=B=C=0;        for(int i=0;i<strlen(g)-1;i++)            A=A*10+g[i]-'0';        sf("%s",g);        sf("%s",g);        for(int i=0;i<strlen(g)-1;i++)            B=B*10+g[i]-'0';        sf("%s",g);        sf("%s",g);        for(int i=0;i<strlen(g);i++)            C=C*10+g[i]-'0';        if(!A)A=1;        if(!B)B=1;        D=e_gcd(A,B,x,y);        if(C%D)        {            puts("No.\n");            continue;        }        x=x*C/D;//计算一个x        ll t=B/D;//x的周期        x=(x%t+t)%t;//计算一个最小非负整数x        ll k=(C-A*x)/B;//计算对应y        if(k>=0)tag=1;//判断y是否大于0        y=y*C/D;        t=A/D;        y=(y%t+t)%t;        k=(C-B*y)/A;        if(k>=0)tag=1;        puts(tag?"Yes.\n":"No.\n");    }    return 0;}