生理周期

http://poj.org/problem?id=1006

中国剩余定理模板题

#include<stdio.h>
#include<string.h>
int p,e,i,d;
int a[5],m[5];
int item = 1;
int extend_gcd(int a,int b,int &x,int &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    else
    {
        int r = extend_gcd(b,a%b,x,y);
        int t = x;
        x = y;
        y = t-a/b*y;
        return r;
    }
}

void CRT()
{
    int M = 1,tm,ret = 0;
    for(int i = 0; i < 3; i++)
        M*=m[i];
    for(int i = 0; i < 3; i++)
    {
        int x,y;
        tm = M/m[i];
        extend_gcd(tm,m[i],x,y);
        ret = (ret + tm*x*a[i])%M;
    }
    if(ret - d <= 0)
        printf("Case %d: the next triple peak occurs in %d days.
",item,ret+M-d);
    else printf("Case %d: the next triple peak occurs in %d days.
",item,(ret+M)%M-d);
}

int main()
{

    while(~scanf("%d %d %d %d",&p,&e,&i,&d))
    {
        if(p == -1 && e == -1 && i == -1 && d == -1)
            break;
        a[0] = p;
        a[1] = e;
        a[2] = i;
        m[0] = 23;
        m[1] = 28;
        m[2] = 33;
        CRT();
        item++;

    }
    return 0;
}

中国剩余定理

设m1,m2,...,mk两两互质,则下面的同余方程组：
  x = a1 (mod m1)
  x = ak (mod mk)
    .
    .
    .
  x = ak (mod mk)

在0<=x<M = m1*m2*...*mk内有唯一解。
记Mi = M / mi (1<=i<=k) ,因为(Mi,mi)=1,故有两整数pi，qi满足Mi*pi + mi*qi = 1，
如果记ei = Mi*pi ，那么有 
ei = 0 (mod mj) , j<>i
   = 1 (mod mj) , j==i
   
很显然，e1*a1 + e2*a2 + ... + ek*ak 就时方程组的一个解 ，
这个解%M后就得到了最小的非负整数解；

求ei可以用扩展欧几里得来求；