HDU 1573 CRT

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CRT模板题

/** @Date    : 2017-09-15 13:52:21
  * @FileName: HDU 1573 CRT EXGCD.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

LL mod;
LL a[11];
LL b[11];
LL exgcd(LL a, LL b, LL &x, LL &y)
{
	LL d = a;
	if(b == 0)
		x = 1, y = 0;
	else 
	{
		d = exgcd(b, a % b, y, x);
		y -= (a / b) * x;
	}
	return d;
}
LL md(LL x,LL y)
{
	LL res = x % y;
	if(res <= 0)
		res = res + y;
	return res;
}
int main()
{
	int T;
	cin >> T;
	while(T--)
	{
		LL n, m;
		scanf("%lld%lld", &n, &m);
		for(int i = 0; i < m; i++) scanf("%d", a + i);
		for(int j = 0; j < m; j++) scanf("%d", b + j);
		LL rem, mod;
		LL x, y;
		LL ans = 0;
		int flag = 0;
		rem = b[0], mod = a[0];
		for(int i = 1; i < m ; i++)
		{
			LL c = b[i] - rem;
			LL g = __gcd(a[i], mod);
			if(c % g!= 0)
				flag = 1;
			else
			{
				exgcd(mod, a[i], x, y);
				LL tmp = a[i] / g;
				//x = (c / g * x % tmp + tmp) % tmp;// m1x1+m2x2 = c
				x = md(c / g * x, tmp);//x =  
				rem = md(rem + mod * x, mod / g * a[i]);
				//rem = mod * x + rem;//ri + mixi
				mod = mod / __gcd(mod, a[i]) * a[i];//LCM(m1, m2);
				//y mod lcm(m1,m2) = x
			}
		}
		//cout << rem <<"~" << mod << endl;
		if(flag || n < rem)
			printf("0\n");
		else printf("%lld\n", (n - rem) / mod + 1);
	}
    return 0;
}//shi·ne

HDU 1573 CRT

标签：using   win   int   com   res   gmail   exgcd   namespace   class   

原文地址：http://www.cnblogs.com/Yumesenya/p/7553931.html