Poj--1061（拓展欧几里得）
2014-12-12 12:14:58
思路：嘛，一道exgcd水题，列下方程：k*(m - n) + p*L = y - x，以k为a，L为b，（m-n）为x，p为y，y-x为c，建立：ax+by=c，若gcd(m-n,L)｜ｃ则有解，否则无解。最后算出解要转化为最小正整数（形式：x0 + t * (b / gcd(m-n,L))）
1 /************************************************************************* 2 > File Name: 1061.cpp 3 > Author: Natureal 4 > Mail: 564374850@qq.com 5 > Created Time: Fri 12 Dec 2014 11:56:40 AM CST 6 ************************************************************************/ 7 8 #include <cstdio> 9 #include <cstring> 10 #include <cstdlib> 11 #include <cmath> 12 #include <vector> 13 #include <map> 14 #include <set> 15 #include <stack> 16 #include <queue> 17 #include <iostream> 18 #include <algorithm> 19 using namespace std; 20 #define lp (p << 1) 21 #define rp (p << 1|1) 22 #define getmid(l,r) (l + (r - l) / 2) 23 #define MP(a,b) make_pair(a,b) 24 typedef long long ll; 25 const int INF = 1 << 30; 26 27 ll X,Y,M,N,L; 28 29 ll Ex_gcd(ll a,ll b,ll &x,ll &y){ 30 if(b == 0){ 31 x = 1; 32 y = 0; 33 return a; 34 } 35 ll r = Ex_gcd(b,a % b,x,y); 36 ll t = x; 37 x = y; 38 y = t - a / b * y; 39 return r; 40 } 41 42 int main(){ 43 while(scanf("%I64d%I64d%I64d%I64d%I64d",&X,&Y,&M,&N,&L) != EOF){ 44 ll a = M - N,b = L,c = Y - X,d,x,y; 45 d = Ex_gcd(a,b,x,y); 46 if(c % d) 47 printf("Impossible\n"); 48 else{ 49 x *= c / d; 50 ll k = L / d; 51 if(k < 0) k = -k; 52 x = (x % k + k) % k; 53 printf("%I64d\n",x); 54 } 55 } 56 return 0; 57 }
