hdu2669

maksyuki 发表于 oj 分类，标签: 数论-乘法逆元

15 9月 2015

Romantic

Problem Description

The Sky is Sprite.
The Birds is Fly in the Sky.
The Wind is Wonderful.
Blew Throw the Trees
Trees are Shaking, Leaves are Falling.
Lovers Walk passing, and so are You.
................................Write in English class by yifenfei

Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead.

Input

The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)

Output

output nonnegative integer X and integer Y, if there are more answers than the X smaller one will be choosed. If no answer put "sorry" instead.

Sample Input

77 51

10 44

34 79

Sample Output

2 -3

sorry

7 -3

Author

yifenfei

Source

HDU女生专场公开赛——谁说女子不如男

题目类型：乘法逆元

算法分析：直接使用扩展欧几里得求解即可

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-6;
const double PI = 2 * acos (0.0);
const int maxn = 1e6 + 66;


LL ex_gcd ( LL a, LL b, LL &x, LL &y)
{
    if (b == 0) {x = 1; y = 0; return a;}

    LL d = ex_gcd (b, a % b, y, x);
    y = y - a / b * x;
    return d;
}

LL Sol (LL a, LL b, LL c)
{
    LL x, y, d;
    d = ex_gcd (a, b, x, y);
    if (c % d)
        return -1;

    x = x * (c / d) % (b / d);
    if (x < 0)
        x += b / d;

    return x;
}


int main()
{
//    CPPFF;
    LL a, b;
    while (cin >> a >> b)
    {
        LL tt = Sol (a, b, 1);
        if (tt == -1)
            cout << "sorry" << endl;

        else
        {
            LL aa = (1 - a * tt) / b;
            cout << tt << " " << aa << endl;
        }
    }
    return 0;
}

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-6;

const double PI = 2 * acos (0.0);

const int maxn = 1e6 + 66;

LL ex_gcd ( LL a, LL b, LL &x, LL &y)

{

if (b == 0) {x = 1; y = 0; return a;}

LL d = ex_gcd (b, a % b, y, x);

y = y - a / b * x;

return d;

}

LL Sol (LL a, LL b, LL c)

{

LL x, y, d;

d = ex_gcd (a, b, x, y);

if (c % d)

return -1;

x = x * (c / d) % (b / d);

if (x < 0)

x += b / d;

return x;

}

int main()

{

// CPPFF;

LL a, b;

while (cin >> a >> b)

{

LL tt = Sol (a, b, 1);

if (tt == -1)

cout << "sorry" << endl;

else

{

LL aa = (1 - a * tt) / b;

cout << tt << " " << aa << endl;

}

return 0;

}

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