扩展欧几里得--hdu3270

hdu3270

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.”

题意：x，y为非负整数，若存在，输出Yes，否则为No

#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <iomanip>
#include <cstdio>

using namespace std;
typedef long long LL;
typedef pair<double, double> PDD;
typedef pair<LL, LL> PLL;

const LL N = 1e6+50;
const LL MOD = 1e9+7;
const LL INF = 0x3f3f3f3f;

#define lson l, m, rt>>1
#define rson m+1, r, rt>>1|1

LL exgcd(LL a, LL b, LL &x, LL &y)
{
    if(!b)
    {
        x = 1;y = 0;
        return a;
    }
    LL ret = exgcd(b, a%b, y, x);
    y -= x*(a/b);
    return ret;
}

int main()
{
    char sa[100], sb[100], sc[100];
    LL a, b, c, x, y;
    while(scanf("%s + %s = %s", sa, sb, sc) != EOF)
    {
        a = 0, b = 0, c = 0;
        for(int i = 0;sa[i] != 'x';++i)
            a = a*10+sa[i]-'0';
        for(int i = 0;sb[i] != 'y';++i)
            b = b*10+sb[i]-'0';
        for(int i = 0;sc[i];++i)
            c = c*10+sc[i]-'0';
        if(!a) a = 1;
        if(!b) b = 1;
        LL g = exgcd(a, b, x, y);
        if(c%g) puts("No.
");
        else if(!c) puts("Yes.
");
        else
        {
            x *= c/g;y *= c/g;
            LL t = b/g;//x最小加的数
            x = (x%t+t)%t;//x最小的时候
            if((c-a*x)/b > 0)
            {
                puts("Yes.
");continue;
            }
            t = a/g;
            y = (y%t+t)%t;//y最小的时候
            if((c-b*y)/a > 0)
            {
                puts("Yes.
");continue;
            }
            puts("No.
");
        }
    }
    return 0;
}