Strange Way to Express Integers—–不互質的中國剩餘定理

參考文章地址：http://blog.csdn.net/hqd_acm/article/details/6173859

題目：http://poj.org/problem?id=2891

思路：

x=r1 (mod a1)  x=r2 (mod a2)--->a1*x+a2*y=gcd(a1,a2)=d--->r1+a1*x=r2+a2*y  ---> a1*x+a2*y=r2-r1   (1)則(r2-r1)%d!=0,方程(1)的解不存在。否則{    t=a2/d;    x=((x*(r2-r1)/d)%t+t)%t//x的最小整數解    r1=x*a1+r1   //求出原方程X=a1*x+r1    a1=a1*a2/d;      r1=(r1%a1+a1)%a1;  //r1<a1}思想就是要將兩個方程式合并，合并式X=a3x+r3,既要mod a1=r1,而且要mod  a2=r2,所以有a3是a1與a2的最小公倍數，r3是方程a1*x+a2*y=r2-r1的解。

原始碼：

#include <stdio.h>typedef long long ll;ll n,a[100000],r[100000];ll extend_gcd(ll a,ll b,ll &x,ll &y){    ll t,ans;    if(b==0)  {  x=1; y=0; return a ;}    else    {        ans=extend_gcd(b,a%b,x,y);        t=x;   x=y;    y=t-a/b*y;    }    return ans;}void solve(ll a[],ll r[],ll n){    ll a1,a2,r1,r2,x,y,d,t;    int flag;    a1=a[0];  r1=r[0];    if(n==1)    {        if(a1<=r1)  printf("-1\n");    //    else        printf("%lld\n",r1);        return;    }    flag=1;    for(int i=1;i<n;i++)    {        a2=a[i];  r2=r[i];       if(a2<=r2) {  printf("-1\n"); flag=0;  break; }        d=extend_gcd(a1,a2,x,y);        if((r2-r1)%d!=0)  {  printf("-1\n"); flag=0;  break; }        else        {            t=a2/d;            x=(x*(r2-r1)/d%t+t)%t;            r1=a1*x+r1;            a1=a1*a2/d;            r1=(r1%a1+a1)%a1;        }    }    if(flag)  printf("%lld\n",r1);}int main(){    while(scanf("%lld",&n)!=EOF)    {        for(int i=0;i<n;i++)        scanf("%lld %lld",&a[i],&r[i]);        solve(a,r,n);    }}