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POJ 1006-Biorhythms(中國剩余定理)

編輯：C++入門知識

POJ 1006-Biorhythms(中國剩余定理)

Biorhythms Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 113666 Accepted: 35658

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.

古人很厲害。。

題意 ：每個人都有3種周期 每個周期分別為23 28 33天，現在給定每個周期的峰值和一個起始天數，求下一次三個峰值同時出現的時刻。

中國剩余定理：已知某個數x x%3==2 x%5==3 x%7==2 求符合條件的min_x;

已知 兩條定理：定理① 由a%b=c 得(a+k*b)%b=c;(k!=0)

            定理② 由a%b=c 得 （a*k）%b=(a%b+a%b+..a%b)%b=(c*k)%b;

由題意可得 題目要求一個最小整數ans滿足 ans%23==p ans%28==e ans%33==i

設 ans1滿足 ans1%23==p 且 ans1是28和33的公倍數。

設 ans2滿足 ans2%28==e 且 ans2是23和33的公倍數。

設 ans3滿足 ans3%33==i 且 ans3是23和28的公倍數。
然後根據定理2就可以算出來 ans1 ans2 ans3
ans=(ans1+ans2+ans3)%(23*28*33);
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define LL long long
int main()
{
	int p,e,i,d,a,cas=1;
	while(cin>>p>>e>>i>>d)
	{
		if(p==-1&&e==-1&&i==-1&&d==-1) break;
		a=1;
		while(33*28*a%23!=1)
			a++;
		int r1=33*28*a*p;
		a=1;
		while(23*33*a%28!=1)
			a++;
		int r2=23*33*a*e;
		a=1;
		while(23*28*a%33!=1)
			a++;
		int r3=23*28*a*i;
		int ans=(r1+r2+r3)%21252;
		if(ans-d<=0)
			ans+=21252;
		printf("Case %d: the next triple peak occurs in %d days.\n",cas++,ans-d);
	}
	return 0;
}