Solution to a problem P4777 [[template] to expand the Chinese Remainder Theorem (EXCRT)]

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Congruence solving equations (Extended Chinese remainder theorem)

P4777 [template] to expand the Chinese Remainder Theorem (EXCRT)

• To solve congruence equation as follows:

  \(\begin{cases}x \equiv a_1\pmod {b_1}\\x \equiv a_2\pmod {b_2}\\......\\x \equiv a_n\pmod {b_n}\\\end{cases}\)

  Wherein, \ (a_i, B_i \) non-negative integer, \ (B_1, B_2, ..., B_n \) necessarily coprime

• Solving:

  Assumed to have been obtained before the \ (k-1 \) Solutions of equations \ (x_ {k-1}
  \) disposed \ (M = I = {LCM_. 1. 1} ^ {K}-BI \) , i.e. \ (M \) is a front \ (k-1 \) a modulus \ (B \) the least common multiple

  Then:
  For the first \ (k-1 \) equations are satisfied \ (x_ {k-1}
  + tM \ equiv a_i \ pmod {b_i} \ \ (t \ in Z) \) ie: before \ (K -1 \) equations, general solution \ (x_ {k-1} + tM \ \ (t \ in Z) \)

  Desire to obtain a first \ (K \) solution of the equations, and the obtained solution, but also meet the first \ (k-1 \) equations

  It:
  is necessary to first \ (K \) a solution of the equation, the former \ (k-1 \) while passing the solutions of the equations also satisfies the first \ (K \) equations conditions.

  Provided: a first \ (K \) Solutions of equations \ (x_k = x_ {k- 1} + tM \ \ (t \ in Z) \)

  Substituting this solution into the first \ (K \) equations may be obtained:
  \ (X_ {K}. 1-tM + \ equiv a_k \ PMOD b_k} {\)
  namely: \ (tM \ - K-equiv a_k-X_ { 1} \ pmod {b_k} \
  ) where: \ (M, a_k, K-X_ {}. 1, b_k \) are known.

  Use \ (exgcd \) solved this congruence equation can be obtained \ (T \) values.

  The \ (T \) values are substituted back \ (K-x_k = X_ {} + tM. 1 \ \ (T \ in the Z) \) , can be obtained \ (x_k \) values

  For \ (K \) after the time of the operation, the solution of equations can be obtained.

• This question pit:

  data range:

  If you just open \ (long \ long \) , then when I get blown up anyway

  In order to avoid blasting variables, use a technique called rapid multiplication modulo odd \ ((ba) \) wonderful \ ((ka) \) algorithm.

  • Rapid multiplication modulo (Also known as turtle speed multiplier)

    Its essence and power quickly to take more than similar, are binary split applications.

    Suppose \ (K \) a \ (A \) multiplying the
    fast by its decomposed into \ (a \ times 2a \ times 4a \ times .... \) form.
    This can take the edge side take the remainder to prevent bursting variables, data overflow.

    But the shift multiplication time complexity,
    abruptly pulled \ (O (logn) \) level

    ll mul(ll A,ll B,ll mod) //快速乘取余 模板
    {
       ll ans=0;
       while(B>0)
         {
           if(B & 1) ans=(ans+A%mod)%mod;
           A=(A+A)%mod;
           B>>=1;
         }
       return ans;
     }

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On the code:

#include<cstdio>
using namespace std;
typedef long long ll;
ll n;
ll a[100010],b[100010]; 
ll mul(ll A,ll B,ll mod) //快速乘取余 模板
{
    ll ans=0;
    while(B>0)
      {
        if(B & 1) ans=(ans+A%mod)%mod;
        A=(A+A)%mod;
        B>>=1;
      }
    return ans;
}
ll exgcd(ll A,ll B,ll &x,ll &y) //扩展欧几里得 模板
{
    if(!B)
      {
        x=1,y=0;
        return A;
      }
    ll d=exgcd(B,A%B,x,y);
    ll tmp=x;
    x=y , y=tmp-A/B*y;
    return d;
}
ll lcm(ll A,ll B) //求最小公倍数
{
    ll xxx,yyy;
    ll g=exgcd(A,B,xxx,yyy);
    return (A/g*B);
}
ll excrt() //重点:求解同余方程组
{
    ll x,y;
    ll M=b[1],ans=a[1]; //赋初值 
    //M为前k-1个数的最小公倍数，ans为前k-1个方程的通解
    for(int i=2;i<=n;i++)
      {
        ll A=M,B=b[i];
        ll C=(a[i]-ans%B+B)%B; //代表同余方程 ax≡c(mod b) 中a,b,c
            
        ll g=exgcd(A,B,x,y);
        //求得A,B的最大公约数，与同余方程ax≡gcd(a,b)(mod b)的解，
            
        if(C%g) return -1; //无解的情况
            
        x=mul(x,C/g,B); //求得x的值,x即t 
        ans+=x*M;  //获得前k个方程的通解
        M=lcm(M,B); //更改M的值
        ans=(ans%M+M)%M;
      }
    return ans;
}
int main()
{
    scanf("%lld",&n);
    for(int i=1;i<=n;i++)
      scanf("%lld%lld",&b[i],&a[i]);
    ll ans=excrt();
    printf("%lld",ans);
}