Strange Way to Express Integers（扩展欧几里得+乘法逆元+中国剩余定理求解非互质的模线性方程组）

2
8 7
11 9

31

/** 
中国剩余定理（不互质） 
*/  
#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)//ax+by=gcd(a,b)  
{  
    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)//len为模方程的式子总数，a[i]、n[i]分别表示第i个式子的余数、除数  
{  
    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;//无解输出-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;  
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#define PI acos(-1.0)
#define LINF 1000000000000000000LL
#define eps 1e-8
#define LL long long
#define MAXN 100010 
#define MOD 1000000007
using namespace std;
const int INF=0x3f3f3f3f;
LL Mod;  
LL gcd(LL a, LL b)  
{  
    if(b==0)  
        return a;  
    return gcd(b,a%b);  
}  
LL exgcd(LL A,LL &x,LL B,LL &y)
{
    LL x1,y1,x0,y0;
    x0=1;y0=0;
    x1=0;y1=1;
    LL r=(A%B+B)%B;
    LL q=(A-r)/B;
    x=0;y=1;
    while(r)
    {
        x=x0-q*x1;
        y=y0-q*y1;
        x0=x1;
        y0=y1;
        x1=x;y1=y;
        A=B;B=r;r=A%B;
        q=(A-r)/B;
    }
    return B;
}
LL ABS(LL x)
{
	return x>=0?x:-x;
}
/*接下来介绍乘法逆元：
若m≥1，gcd（a，m）＝1，则存在c使得 ca≡1（mod m） 我们把c称为是a对模m的逆，记为 a－1（mod m）或a－1 可以用扩展欧几里德算法求a－1
应用： 求(a/b)%c时，若a为大整数时可以写成 ((a%c)*((b－1)%c))%c,注意：这里(b－1)是b%c的乘法逆元，不是b减 1！！！ 
还可以使用扩展欧几里德算法求乘法逆元：*/
//a在模n乘法下的逆元，没有则返回-1(a * b % n == 1,已知a,n,求b就是求a模n的乘法逆元)    
LL inv(LL a, LL n)  
{  
    LL x,y;  
    //int64 t = Extend_Euclid(a,n,x,y); 
	/*******************************************
	扩展欧几里得算法求乘法逆元
	调用：exgcd(a,x,b,y);
	则：  x为a模b的乘法逆元，y为b模a的乘法逆元
	********************************************/ 
    LL t = exgcd(a,x,n,y);  
    if(t != 1)  
        return -1;  
    return (x%n+n)%n;  
}  
  
//将两个方程合并为一个  
bool merge(LL a1, LL n1, LL a2, LL n2, LL& a3, LL& n3)  
{  
    LL d = gcd(n1,n2);  
    LL 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  
LL China_Reminder2(LL len, LL* a, LL* n)//len为模方程的式子总数，a[i]、n[i]分别表示第i个式子的余数、除数  
{  
    LL a1=a[0],n1=n[0];  
    LL a2,n2;  
    for(int i = 1; i < len; i++)  
    {  
        LL aa,nn;  
        a2 = a[i],n2=n[i];  
        if(!merge(a1,n1,a2,n2,aa,nn))  
            return -1;//无解输出-1  
        a1 = aa;  
        n1 = nn;  
    }  
    Mod = n1;  
    return (a1%n1+n1)%n1;  
}  
LL 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;  
}