[POJ] #1006# Biorhythms : 最小公倍数/同余问题

一. 题目

Biorhythms

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 127263		Accepted: 40329

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.

Source

East Central North America 1999

 

二. 题意

• 每个人一年分别由三个不同的巅峰日，周期分别为23， 28， 33
• 给出当年三个巅峰日: p, e, i, 一个指定开始日: d
• 求出从d日开始，经历多少天之后，下次三个巅峰日重合

 

三. 分析

• 算法核心: 最小公倍数/同余问题
• 实现细节:
  • 题目给出了条件，假设下次三个巅峰重合日小与等于21252
  • 21252实际上是23, 28, 33的最小公倍数
  • 对于此题，实际上就是求出d日后的某一天 j（<=21252），使得
    
    • (j - p) % 23 == 0
    • (j - e) % 28 == 0
    • (j - i) % 33 == 0

 

四. 题解

#include <stdio.h>
#define CYCLE_P 23
#define CYCLE_E 28
#define CYCLE_I 33
#define MAX_DAY 21252

int main()
{
    int p, e, i, d, count = 0;

    while (1) {
        int j;
        count++;

        scanf("%d %d %d %d\n", &p, &e, &i, &d); 
        
        if (p == -1 && e == -1 && d == -1) break;

        for (j = d + 1; j < MAX_DAY; j++)
            if (0 == (j - p) % CYCLE_P) break;

        for (; j < MAX_DAY; j += CYCLE_P)
            if (0 == (j - e) % CYCLE_E) break;

        for (; j < MAX_DAY; j += CYCLE_P * CYCLE_E)
            if (0 == (j - i) % CYCLE_I) break;

        printf("Case %d: the next triple peak occurs in %d days.\n",count,j - d);
    }

    return 0;
}