原文博客: http://blog.csdn.net/shuangde800/article/details/8236891
鏈接:
http://codeforces.com/problemset/problem/7/C
題目大意:
給方程Ax + By + C = 0. 其中A,B,C爲已知, 求x,y。
分析與總結:
拓展歐幾里得算法的模板題。這個算法在數論書或者網上都可以找到。
該算法求出線性方程Ax + By = gcd(A, B);
然後,這個方程可進行轉換:
Ax + By = gcd(A, B)
=> Ax + By = -C/z, 其中-C/z = gcd(A, B)
=> Ax*z + By*z = C.
其中x, y可以通過拓展歐幾里得算法求出,
然後,我們只需要求出z, 而z = -C/gcd(A,B);
所以, 最終答案x = x*(-C/gcd(A,B)) , y = y*(-C/gcd(A,B));
代碼:
- #include<iostream>
- #include<cstdio>
- #include<cstring>
- using namespace std;
- typedef long long LL;
- const LL INF = 5*1e18;
- void gcd(LL a, LL b, LL& d,LL& x, LL& y){
- if(!b){d=a; x=1; y=0; }
- else {gcd(b,a%b,d,y,x); y -= x*(a/b); }
- }
- int main(){
- LL a,b,c,d,x,y;
- cin >> a >> b >> c;
- gcd(a,b,d,x,y);
- if(c%d != 0)
- puts("-1");
- else
- cout << -x*(c/d) << " " << -y*(c/d) << endl;
- return 0;
- }
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const LL INF = 5*1e18;
void gcd(LL a, LL b, LL& d,LL& x, LL& y){
if(!b){d=a; x=1; y=0; }
else {gcd(b,a%b,d,y,x); y -= x*(a/b); }
}
int main(){
LL a,b,c,d,x,y;
cin >> a >> b >> c;
gcd(a,b,d,x,y);
if(c%d != 0)
puts("-1");
else
cout << -x*(c/d) << " " << -y*(c/d) << endl;
return 0;
}