中国剩余定理 poj 1006 poj2891

Biorhythms

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 129686		Accepted: 41277

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.

题意：

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

题解：使用中国剩余定理

（互质的模线性方程组）

中国剩余定理（CRT）的表述如下

 

设正整数两两互素，则同余方程组

 

                             

 

有整数解。并且在模下的解是唯一的，解为

 

                               

 

其中，而为模的逆元。

逆元概念：

群G中任意一个元素a，都在G中有唯一的逆元a‘,具有性质aa'=a'a=e，其中e为群的单位元

例如：4关于1模7的乘法逆元为多少？

4X≡1 mod 7

这个方程等价于求一个X和K，满足

4X=7K+1

其中X和K都是整数。

若ax≡1 mod f, 则称a关于模f的乘法逆元为x。也可表示为ax≡1(mod f)。

当a与f互素时，a关于模f的乘法逆元有解。如果不互素，则无解。如果f为素数，则从1到f-1的任意数都与f互素，即在1到f-1之间都恰好有一个关于模f的乘法逆元。

例如，求5关于模14的乘法逆元：

14=5*2+4

5=4*1+1

说明5与14互素，存在5关于14的乘法逆元。

1=5-4=5-(14-5*2)=5*3-14

因此，5关于模14的乘法逆元为3。

其求法可用欧几里得算法，可以见我的另外一片blog， 详细推导扩展欧几里得算法

http://blog.csdn.net/summer__show_/article/details/52238225

Extended Euclid (d，f) //算法求d关于模f的乘法逆元d -1 ，即 d* d -1 mod f = 1

#include
#include
#include
using namespace std;

void ex_gcd(int a1,int b,int &x,int &y)
{
    if(b==0){
        x=1,y=0;
        return ;
    }
    ex_gcd(b,a1%b,x,y);
    int temp=x;
    x=y;
    y=temp-(a1/b)*y;
}

int CRT(int a[],int m[],int n)
{
    int M=1,ans=0;
    for(int i=0;i

下面是中国剩余定理的特殊情况：

不互质的模线性方程组

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 14111		Accepted: 4564

Description

Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:

Choose k different positive integers a₁, a₂, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_k are properly chosen, m can be determined, then the pairs (a_i, r_i) can be used to express m.

“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”

Since Elina is new to programming, this problem is too difficult for her. Can you help her?

Input

The input contains multiple test cases. Each test cases consists of some lines.

• Line 1: Contains the integer k.
• Lines 2 ~ k + 1: Each contains a pair of integers a_i, r_i (1 ≤ i ≤ k).

Output

Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.

Sample Input

2
8 7
11 9

Sample Output

31

Hint

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

题意：

就是给你n组数据

每一组数据a，b   表示    x%a==b

求这n组数据的公共的最小x的解

题解：

中国剩余定理中的不互质的模线性方程组

由不互质的模线性方程组
x%r1=a1              1
x%r2=a2              2
x%r3=a3              3
x%r4=a4              4
......
由  1  2式子得到：
x = r1*k1 + a1
x = r2*k2 + a2
联立：
r1*k1 + a1 =r2*k2 + a2
得到
r1*k1 + r2*k2 = a2 - a1  （其中k2的正负可变）

令g=gcd（r1，r2）

那么上叙方程同时除以g后

r1*k1 / g + r2*k2 / g = （a2 - a1 ）/ g            

由扩展欧几里得可以得到k1，则有
x1 = r1*k1 + a1                         5
     x1表示1 2式子得到的r1 和 让r2的最大公约数

（扩展欧几里得算法详解http://blog.csdn.net/summer__show_/article/details/52238225）

因此1 2式子可以合并成
x = t*lcm(r1,r2) + x1                        
此时 x 是r1   r2  的最小公倍数，可以返回去验证一下即可


在程序中把5式子变成x = r1*k1 + a1
即 k1=t   r1=gcd(r1,r2)   a1=x1  

然后一直循环求解下去

#include
#include
#include
using namespace std;


#define LL long long


LL ex_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0){
        x=1,y=0;
        return a;
    }


    LL d=ex_gcd(b,a%b,x,y);
    LL temp=x;
    x=y;
    y=temp-(a/b)*y;
    return d;
}


int main()
{
    LL k,a1,a2,r1,r2;
    LL x,y,d;
   // freopen("in.txt","r",stdin);
    while(scanf("%I64d",&k)!=EOF)
    {
        scanf("%I64d%I64d",&a1,&r1);
        k--;
        int flag=0;
        while(k--)
        {
            scanf("%I64d%I64d",&a2,&r2);
            if(flag)
                continue;


            //扩展欧几里得
            LL a=a1,b=a2;
            LL c=r2-r1;
            d=ex_gcd(a,b,x,y);//最小公约数
            if(c%d){
                flag=1;
                continue;
            }
            
            LL t=b/d;//除以d表示方程两边都除以d
            x=(x*(c/d)%t+t)%t;
            r1=a1*x+r1;//同上面所说的式子5
            a1=a1*(a2/d);//a1 a2最大公倍数
        }


        if(flag)
            printf("-1\n");
        else
            printf("%I64d\n",r1);
    }
    return 0;
}