hdu1573 X问题

原题：
http://acm.hdu.edu.cn/showproblem.php?pid=1573
这个题目真是坑爹啊- =首先说M个正整数，结果测试数据里面就有0 - =莫非0成了正整数？
扩展中国剩余定理解这道题目很多模板也不对- =造呢。。也不知道到底可不可以从0开始- =坑爹的题目。。

如果时间急别在这里干着急了- =坑死人

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define lln long long int

lln a[11];
lln b[11];

lln gcd( lln a, lln b)
{
return b == 0 ? a : gcd(b, a % b);
}

void ace()
{
int n, m;
int t;
int i, j;
int lcm;
bool flag;
int num;
cin >> t;
while (t--)
{
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
cin >> n >> m;
lcm = 1;
flag = false;
num = 0;
for (i = 0; i < m; i++)
{
cin >> a[i];
lcm = lcm * a[i] / gcd(lcm, a[i]);
}
for (i = 0; i < m; i++)
cin >> b[i];
for (i = 1; i <= lcm && i <= n; i++) //find the minium num 坑爹的地方- - i不能从0开始
{
for (j = 0; j < m; j++)
if (i % a[j] != b[j])
break;
if (j == m)
{
flag = true;
break;
}
}
if (flag)
{
int temp = n % lcm;
if (temp >= i)
i = n / lcm + 1;
else
i = n / lcm;
printf("%d\n", i);
//			while (i <= n)
//			{
//				i += lcm;
//				num++;
//			}
//			cout << num << endl;
}
else
printf("0\n");
}
}
int main()
{
ace();
return 0;
}

/**
中国剩余定理（不互质）模板
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef __int64 int64;
int64 Mod;

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;
}

//a在模n乘法下的逆元，没有则返回-1
int64 inv(int64 a, int64 n)
{
int64 x,y;
int64 t = Extend_Euclid(a,n,x,y);
if(t != 1)
return -1;
return (x%n+n)%n;
}

//将两个方程合并为一个
bool merge(int64 a1, int64 n1, int64 a2, int64 n2, int64& a3, int64& n3)
{
int64 d = gcd(n1,n2);
int64 c = a2-a1;
if(c%d)
return false;
c = (c%n2+n2)%n2;
c /= d;
n1 /= d;
n2 /= d;
c *= inv(n1,n2);
c %= n2;
c *= n1*d;
c += a1;
n3 = n1*n2*d;
a3 = (c%n3+n3)%n3;
return true;
}

//求模线性方程组x=ai(mod ni),ni可以不互质
int64 China_Reminder2(int len, int64* a, int64* n)
{
int64 a1=a[0],n1=n[0];
int64 a2,n2;
for(int i = 1; i < len; i++)
{
int64 aa,nn;
a2 = a[i],n2=n[i];
if(!merge(a1,n1,a2,n2,aa,nn))
return -1;
a1 = aa;
n1 = nn;
}
Mod = n1;
return (a1%n1+n1)%n1;
}
int64 a[1000],b[1000];
int main()
{
int i;
int k;
while(scanf("%d",&k)!=EOF)
{
for(i = 0; i < k; i++)
scanf("%I64d %I64d",&a[i],&b[i]);
printf("%I64d\n",China_Reminder2(k,b,a));
}
return 0;
}