X问题

题意:

题意就是让你求中国剩余定理解的个数..第一次接触CRT,一开始的思路是先求出来a1到an的lcm，再用CRT求出来方程的一个解x，然后用这个x逐个加lcm去扩充。。然后用搞了半天WA了一发..然后借助着网上的模版A了...稍后补篇blog码一下CRT的相关知识....

实际上只是用了CRT的思想,直接求a1~an的LCM 然后在LCM范围内寻找满足题意的解,同时也要保证在n的范围内,如果解X满足题意 那么X+LCM也满足这样.

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;
typedef __int64 int64;
int64 Mod;
long long n;
int M;
int64 gcd(int64 a, int64 b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}

int64 Extend_Euclid(int64 a, int64 b, int64&x, int64& y)
{
    if(b==0)
    {
        x=1,y=0;
        return a;
    }
    int64 d = Extend_Euclid(b,a%b,x,y);
    int64 t = x;
    x = y;
    y = t - a/b*y;
    return d;
}


//求模线性方程组x=ai(mod ni),ni可以不互质
int64 China_Reminder2(int len,int64 w[],int64 Md[])
{
	bool flag=false;
	int64 n1=Md[0], b1=w[0],x,y;
	for (int i=1;i<len;i++)
	{
		int64 n2=Md[i], b2=w[i];
		int64 bb=b2-b1;
		int64 d=Extend_Euclid(n1,n2,x,y);
		if (bb % d)	
		{
			flag = true;
			break;
		}
		int64 k = bb/d*x;
		int64 t = n2/d;
		if (t < 0) t = -t;
		k = (k % t + t) % t;	
		b1 = b1 + n1*k;
		n1 = n1 / d * n2;
	}
	if (flag) return 0; 
    if (b1 == 0)   
        b1 = n1;   
    if (b1 > n)
        return 0;
    return (n-b1)/n1+1;    
}  

int64 a[11],b[11];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%d",&n,&M);
		//cout<<"t="<<t<<" N="<<N<<endl;
		for(int i=0;i<M;i++)scanf("%lld",&a[i]);
		for(int i=0;i<M;i++)scanf("%lld",&b[i]);
		long long res=China_Reminder2(M,b,a);
		printf("%d\n",res);
	}
	return 0;	
}