sincerit 2669 Romantic 擴充套件歐幾里得
2669 Romantic
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10370 Accepted Submission(s): 4427
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 Xa + Yb = 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
ax + by = gcd(a,b)
當gcd(a,b) == 1時 想,即為答案
當gcd(a,b) != 1時, 先當成1計算最後結果x,y同乘與gcd(a,b)
當要求x大於0時, while (x <=0 ) x += b, y -= a;
#include <iostream>
using namespace std;
typedef long long ll;
// ax + by = 1;
ll extgcd(ll a, ll b, ll &x, ll &y) {
ll temp;
if (b == 0) {
x = 1;
y = 0;
return a;
} else {
temp = extgcd(b, a%b, y, x);
y -= (a/b) * x;
}
return temp;
}
int main() {
ll n, m;
while (cin >> n >> m) {
ll x, y;
ll k = extgcd(n, m, x, y);
if (k != 1) cout << "sorry\n";
else {
// a(x+b) + b(y-a) == ab + ax + by - ab = 1
while (x <= 0) {
x += m;
y -= n;
}
cout << x << " " << y << "\n";
}
}
return 0;
}