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
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
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#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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