求解模线性方程组(中国剩余定理)
x=a1 mod m1
x=a2 mod m2
......
x=an mod mn 其中,a[],m[]已知,m[i]>0且m[i]与m[j]互质,求x.
普通的中国剩余定理要求所有的m是互素
有整数解。并且在模下的解是唯一的，解为
int CRT(int a[],int m[],int n)
{
int M = 1;
int ans = 0;
for(int i=1; i<=n; i++)
M *= m[i];
for(int i=1; i<=n; i++)
{
int x, y;
int Mi = M / m[i];
extend_Euclid(Mi, m[i], x, y);
ans = (ans + Mi * x * a[i]) % M;
}
if(ans < 0) ans += M;
return ans;
}
人自出生起就有体力，情感和智力三个生理周期，分别为23，28和33天。一个周期内有一天为峰值，在这一
天，人在对应的方面（体力，情感或智力）表现最好。通常这三个周期的峰值不会是同一天。现在给出三个日
期，分别对应于体力，情感，智力出现峰值的日期。然后再给出一个起始日期，要求从这一天开始，算出最少
再过多少天后三个峰值同时出现。
0 0 0 0
0 0 0 100
5 20 34 325
4 5 6 7
283 102 23 320
203 301 203 40
-1 -1 -1 -1
Case 1: the next triple peak occurs in 21252 days.
Case 2: the next triple peak occurs in 21152 days.
Case 3: the next triple peak occurs in 19575 days.
Case 4: the next triple peak occurs in 16994 days.
Case 5: the next triple peak occurs in 8910 days.
Case 6: the next triple peak occurs in 10789 days.
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int a[4], m[4];
void extend_Euclid(int a, int b, int &x, int &y)
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
extend_Euclid(b, a % b, x, y);
int tmp = x;
x = y;
y = tmp - (a / b) * y;
}
int CRT(int a[],int m[],int n)
{
int M = 1;
int ans = 0;
for(int i=1; i<=n; i++)
M *= m[i];
for(int i=1; i<=n; i++)
{
int x, y;
int Mi = M / m[i];
extend_Euclid(Mi, m[i], x, y);
ans = (ans + Mi * x * a[i]) % M;
}
if(ans < 0) ans += M;
return ans;
}
int main()
{
int p, e, i, d, t = 1;
while(cin>>p>>e>>i>>d)
{
if(p == -1 && e == -1 && i == -1 && d == -1)
break;
a[1] = p;
a[2] = e;
a[3] = i;
m[1] = 23;
m[2] = 28;
m[3] = 33;
int ans = CRT(a, m, 3);
if(ans <= d)
ans += 21252;
cout<<"Case "<<t++<<": the next triple peak occurs in "<<ans - d<<" days."<<endl;
}
return 0;
}
不互素？
input：
2
8 7
11 9
output：
31
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
typedef long long LL;
const int N = 1005;
LL a[N], m[N];
LL gcd(LL a,LL b)
{
return b? gcd(b, a % b) : a;
}
void extend_Euclid(LL a, LL b, LL &x, LL &y)
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
extend_Euclid(b, a % b, x, y);
LL tmp = x;
x = y;
y = tmp - (a / b) * y;
}
LL Inv(LL a, LL b)
{
LL d = gcd(a, b);
if(d != 1) return -1;
LL x, y;
extend_Euclid(a, b, x, y);
return (x % b + b) % b;
}
bool merge(LL a1, LL m1, LL a2, LL m2, LL &a3, LL &m3)
{
LL d = gcd(m1, m2);
LL c = a2 - a1;
if(c % d) return false;
c = (c % m2 + m2) % m2;
m1 /= d;
m2 /= d;
c /= d;
c *= Inv(m1, m2);
c %= m2;
c *= m1 * d;
c += a1;
m3 = m1 * m2 * d;
a3 = (c % m3 + m3) % m3;
return true;
}
LL CRT(LL a[], LL m[], int n)
{
LL a1 = a[1];
LL m1 = m[1];
for(int i=2; i<=n; i++)
{
LL a2 = a[i];
LL m2 = m[i];
LL m3, a3;
if(!merge(a1, m1, a2, m2, a3, m3))
return -1;
a1 = a3;
m1 = m3;
}
return (a1 % m1 + m1) % m1;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=1; i<=n; i++)
scanf("%I64d%I64d",&m[i], &a[i]);
LL ans = CRT(a, m, n);
printf("%I64d\n",ans);
}
return 0;
}
设m1,m2,...,mn是两两互素的正数,则对任意的整数a1,a2,...,an,同余方程组
其解为:X=((M_1*M1*a1)+(M_2*M2*a2)+...+(M_n*Mn*an)) mod m;
其中m=m1*m2*...*mn; Mi=m/mi; M_i是Mi的逆元
int china(int *a,int *m,int n)
{
int M=1,ans=0,mi,i,x,y;
for(i=0;i<n;i++) M*=m[i];
for(i=0;i<n;i++)
{
mi=M/m[i];
exgcd(m[i],mi,x,y);
ans=(ans+mi*y*a[i])%M;
}
return (ans%M+M)%M;
}
