Chinese remainder theorem (solving a linear congruence equation)

problem:

　　There is such a problem in the "Sun Tzu Suan Jing" said: "Today there was an unknown number three number three of the remaining two (remainder divided by 32), fifty-five number of remaining three (3 divided by more than 5) Weeks the number of the remaining two (divided by more than 72), asked the geometric objects? "this problem is called" grandson problem ", called on the international general solution to the problem," Chinese remainder Theorem. "

Resolution:

　　The meaning is the subject of such a set of equations:

(M1 --- mn pairwise mass)

Assume:

Set up: So:

So the equation is solvable:

Find modM meaning under, or the smallest positive integer solution: the equation has a unique solution:

Note: Assuming an integer of m1, m2, m3 ... mn pairwise mass, then for any integer a1, a2, a3 ... an, x solvability

Example:

for

x % 3 = 2

x % 5 = 3

x % 7 = 2

We need to construct an answer // inv is the inverse yuan

5*7*inv(5*7, 3) % 3 = 1

3*7*inv(3*7, 5) % 5 = 1

3*5*inv(3*5, 7) % 7 = 1

Then take sides with the number you need, too

2 * 5*7*inv(5*7, 3) % 3 = 2

3 * 3*7*inv(3*7, 5) % 5 = 3

2 * 3*5*inv(3*5, 7) % 7 = 2

make

a = 2 * 5*7*inv(5*7, 3)

b = 3 * 3*7*inv(3*7, 5)

c = 2 * 3*5*inv(3*5, 7)

Then

a % 3 = 2

b % 5 = 3

c % 7 = 2

The answer is a + b + c

because

a% 5 = a% 7 = 0 // a is a multiple of 5 * 7 * inv (5 * 7,3), so the mold 5 and 7 are

b% 3 = b% 7 = 0 // Similarly, b, c mold 3 / 7,3 / 5 are all 0

c%3 = c%5 = 0

and so

(a + b + c) % 3 = (a % 3) + (b % 3) + (c % 3) = 2 + 0 + 0 = 2

(a + b + c) % 5 = (a % 5) + (b % 5) + (c % 5) = 0 + 3 + 0 = 3

(a + b + c) % 7 = (a % 7) + (b % 7) + (c % 7) = 0 + 0 + 2 = 2

I.e., a + b + c is a number satisfying the condition to give @ answer smallest positive integer solution is obtained (a + b + c)% 105 (105 = 3 * 5 * 7, pairwise nature, the least common multiple of 105)

Code:

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <cstdio>
#include <queue>
#include <cmath>
#include <map>
#include <set>

using namespace std;

typedef long long ll;

const int maxn=1e5+10;

ll a[maxn],m[maxn],n;

ll ex_gcd(ll a,ll b,ll &x,ll &y){
    if(b==0){x=1;y=0;return a;}
    ll ans=ex_gcd(b,a%b,x,y);
    ll temp=x;
    x=y;
    y=temp-a/b*y;
    return ans;
}
ll inv(ll a,ll b){   //求逆
   ll x,y;
   ll ans=ex_gcd(a,b,x,y);
   if(ans!=1)return -1;
   if(x<0)x=(x%b+b)%b;
   return x;
}
ll China(){//中国剩余定理
   ll M=1;
   for(int i = 0;i<n;i++){
      M*=m[i];
   }
   ll sum=0;
   for(int i=0;i<n;i++){
      ll res=M/m[i];
      sum=(sum+a[i]*res*inv(res,m[i]))%M;
   }
   return sum;
}

int main(){
    while(scanf("%lld",&n)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%lld%lld",&m[i],&a[i]);
        }
        ll ans=China();
        printf("%lld\n",ans);
    }
    return 0;
}

The above methods can only solve m twenty-two congruence equation relatively prime, if not relatively prime it?

Act One:

Assuming that only two equations, if they can use an equation instead of two equations, and can get the solution of this equation, then the problem is solved:

x=a1+ k1*b1 (x%b1= a1)

x=a2+ k2*b2 (x%b2 = a2)

Then: a1l + k1 * b1 = a2 + k2 * b2

then: b1* k1 + b2*(-k2)= a2-a1

Suppose gcd (b1, b2) = g if (a2 - a1)% g = 0:! No solution

Otherwise: k1 * b1 = (a2-a1) + k2 * b2

then: k1*b1/g = (a2-a1)/g + k2*b2/g

then: k1*b1/g = (a2-a1)/g (mod b2/g)

then : k1 = inv(b1/g，b2/g)*(a2-a1)/g (mod b2/g)

then:x = a1 + inv(b1/g，b2/g)*(a2-a1)/g*b1 + b2*b1/g*y

then :x = c(mod m)

中:c= a1 + inv(b1/g,b2/g)*(a2-a1)/g*b1

m= b1*b2/g = lcm(b1,b2)

As a new equation simultaneous continue with the next, until the last remaining equation xn = cn (mod mn) then,

A solution of any of the conditions are satisfied xn, can take the smallest positive integer ans = (cn% mn + mn)% mn

$x \equiv a_{1}$

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <cstdio>
#include <queue>
#include <cmath>
#include <map>
#include <set>

using namespace std;

typedef long long ll;
const int maxn=1e5+10;
ll C[maxn],M[maxn];
ll n,k;
ll gcd(ll a,ll b){
    return b==0?a:gcd(b,a%b);
}
ll ex_gcd(ll a,ll b,ll &x,ll &y){
    if(b==0){x=1;y=0;return a;}
    ll g=ex_gcd(b,a%b,x,y);
    ll temp=x;
    x=y;
    y=temp-a/b*y;
    return g;
}
ll inv(ll a,ll mod){
    ll X,Y;
    ll g=ex_gcd(a,mod,X,Y);
    if(g!=1)return -1;
    return (X%mod+mod)%mod;
}
/*
ll mul(ll a,ll b,ll mod){   //快速乘法
    ll ans=0;
    while(b){
        if(b&1)ans=(ans%mod+a%mod)%mod;
        b>>=1;
        a=(a%mod+a%mod)%mod;
    }
    return ans;
}
*/
int main(){
    while(scanf("%lld",&n)!=EOF){
        for(int i = 0;i<n;i++){
            scanf("%lld%lld",&M[i],&C[i]);
        }
        bool flag=true;
        for(int i=1;i<n;i++){
            ll M1=M[i-1],M2=M[i],C1=C[i-1],C2=C[i];
            ll g=gcd(M1,M2);
            if((C2-C1)%g){flag = false;break;}
            M[i]=M1/g*M2;   //可能会爆
            ll INV=inv(M1/g,M2/g);
            if(INV==-1){flag=false;break;}
            C[i]= C1+(INV*((C2-C1)/g))%(M2/g)*M1;
            C[i]=(C[i]%M[i]+M[i])%M[i];
        }
        if(!flag)printf("-1\n");
        else printf("%lld\n",C[n-1]);
    }
    return 0;
}

Act II :( m only a relatively small time / cow off the 2019 summer school more than 1-d)

Note that before the configuration of i equations minimum natural number solution, before assuming the minimum natural number i-1 th equation is f (i-1). lcm (p1, p2, ......, pi-1) = y, so that f (i) = f (i-1) and let f (i) each plus y, the i-th known equations can be satisfied. In the case of guarantee can prove solvability, adding no more than pi times. (It first determines whether or solutions of equations)