POJ 2155 C Looooops

来源：http://poj.org/problem?id=2115

题意：就是说一个for循环，变量从a开始，每次增加c，问经过多少次增加能够 (a+c*n) % 2^k = b%2^k，即同余方程。经过变形后可以变成裸的扩展欧几里得

思路：用扩展欧几里得解就可以了。。

代码：

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;

typedef long long ll;
ll gcd(ll a,ll b){
	if(b == 0)
		return a;
	return gcd(b,a%b);
}
ll mi(ll x){
	ll f = 1;
	for(int i = 1;i <= x; ++i)
		 f*= 2;
	return f;
}
void extend_Eulid(ll a,ll b, ll &x,ll &y){
	if(b == 0){
	  x = 1;
	  y = 0;
	  return;
	}
	else{
	  extend_Eulid(b,a%b,x,y);
	  ll temp = x;
	  x = y;
	  y = temp - a/b * y;
	}
}
int main(){
	ll a,b,c,k;
	while(scanf("%lld%lld%lld%lld",&a,&b,&c,&k)){
	  if(a + b + c + k == 0)
		  break;
	  ll ans = mi(k);
	  ll gcdbc = gcd(ans,c);
	  ll value = (b - a + ans)%ans;
	  if(value % gcdbc)
		  printf("FOREVER\n");
	  else{
	    c /= gcdbc;
		ans /= gcdbc;
		value /= gcdbc;
		ll xx,yy;
		extend_Eulid(ans,c,xx,yy);
		xx *= value;
		yy *= value;
		yy %= ans;
		if(yy < 0){
		  yy = (yy+ans)%ans;
		}
		printf("%lld\n",yy);
	  }
	}
	return 0;
}