【学习笔记-中国剩余定理】POJ1006 Biorhythms

Biorhythms

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 139500		Accepted: 44772

Description

Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical, emotional, and intellectual cycles, and they have periods of lengths 23, 28, and 33 days, respectively. There is one peak in each period of a cycle. At the peak of a cycle, a person performs at his or her best in the corresponding field (physical, emotional or mental). For example, if it is the mental curve, thought processes will be sharper and concentration will be easier. 
Since the three cycles have different periods, the peaks of the three cycles generally occur at different times. We would like to determine when a triple peak occurs (the peaks of all three cycles occur in the same day) for any person. For each cycle, you will be given the number of days from the beginning of the current year at which one of its peaks (not necessarily the first) occurs. You will also be given a date expressed as the number of days from the beginning of the current year. You task is to determine the number of days from the given date to the next triple peak. The given date is not counted. For example, if the given date is 10 and the next triple peak occurs on day 12, the answer is 2, not 3. If a triple peak occurs on the given date, you should give the number of days to the next occurrence of a triple peak. 

Input

You will be given a number of cases. The input for each case consists of one line of four integers p, e, i, and d. The values p, e, and i are the number of days from the beginning of the current year at which the physical, emotional, and intellectual cycles peak, respectively. The value d is the given date and may be smaller than any of p, e, or i. All values are non-negative and at most 365, and you may assume that a triple peak will occur within 21252 days of the given date. The end of input is indicated by a line in which p = e = i = d = -1.

Output

For each test case, print the case number followed by a message indicating the number of days to the next triple peak, in the form:

Case 1: the next triple peak occurs in 1234 days.

Use the plural form ``days'' even if the answer is 1.

Sample Input

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

Sample Output

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.
 
题解：中国剩余定理 参考
设m1,m2,m3,m4两两互素，则同余方程组
x≡a1(m1)
x≡a2(m2)
x≡a3(m3)
x≡a4(m4)
....
x≡ak(mk)
一定有解，x≡(a1*M1*M1^(-1)+a2*M2*M2^(-1)+....)
其中M=m1*m2*...*mk，Mi=M/mi，Mi^(-1)是Mi在模mi意义下的逆元。
普通的中国剩余定理要求所有mi互素，那么如果不互素呢？
我们采用两两合并的思想，假设要合并如下两个方程
x=a1+m1*x1
x=a2+m2*x2
那么得到
a1+m1x1=a2+m2x2 => m1x1+m2x2=a2-a1
再利用扩展欧几里得算法解出x1的最小正整数解，再带入
x=a1+m1x1，得到x后合并为一个方程的结果过为
y≡x(mod lcm(m1,m2))
这样一直合并下去，最终可以求得同余方程的解。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
 
int a[],m[];
int p,e,i,d,t=;
 
void exgcd(int a,int b,int &x,int &y){
    if(b==){
        x=;
        y=;
        return;
    }
    exgcd(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=,ans=;
    for(int i=;i<=n;i++)M*=m[i];
    for(int i=;i<=n;i++){
        int x,y;
        int Mi=M/m[i];
        exgcd(Mi,m[i],x,y);
        ans=(ans+Mi*x*a[i])%M;
    }
    if(ans<)ans+=M;
    return ans;
}
 
int main(){
    while(cin>>p>>e>>i>>d){
        if(p==-&&e==-&&i==-&&d==-)break;
        a[]=p;a[]=e;a[]=i;
        m[]=;m[]=;m[]=;
        int ans=CRT(a,m,);
        if(ans<=d)ans+=;
        printf("Case %d: the next triple peak occurs in %d days.\n",t++,ans-d);
    }
    return ;
}

不互素的

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 5
#define ll long long
using namespace std;
ll n,m[N],a[N],m1,e;
ll read()
{
    ll x=,f=; char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-; ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}
    return x*f;
}
ll exgcd(ll a,ll b,ll &x,ll &y)
{
    if(b==)
    {
        x=,y=;
        return a;
    }
    ll r=exgcd(b,a%b,x,y),tmp;
    tmp=x,x=y,y=tmp-a/b*y;
    return r;
}
ll crt()
{
    ll a1=a[],a2,m2,d,c;m1=m[];
    for(ll i=;i<=n;++i)
    {
        a2=a[i],m2=m[i];
        c=a2-a1;ll x=,y=;
        d=exgcd(m1,m2,x,y);
        if(c%d) return  -;
        x=x*c/d;
        int mod=m2/d;
        x=(mod+x%mod)%mod;
        a1+=m1*x;m1*=mod;
    }
    return a1;
}
int main()
{
//    freopen("mod.in","r",stdin);
//    freopen("mod.out","w",stdout);
    n=;
    for(int i=;i<=n;i++)
     m[i]=read(),a[i]=read();
    printf("%lld\n",crt());
    return ;
}