九度OJ 1142：Biorhythms（生理周期） （中国剩余定理）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：266

解决：189

题目描述：

    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.

输入：

    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. 

输出：

    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. 

样例输入：

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

样例输出：

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.

来源：

2009年北京大学计算机研究生机试真题

思路：

我的思路是，对三个周期，每次最小那个天数的加一个周期，直到三个周期重合。

本题还可用中国剩余定理求解，具体自己上网搜吧，很多相关文章。

九度上能过的在POJ上没过，原来给的起始天数可能已经超过了d时间开始的第一个重合周期了，要将起始天数恢复到d之前才行。

代码1是九度上能过的C代码，代码2是POJ上能过的C++代码。

代码1：

#include <stdio.h>
 
int min(int a[3])
{
    int m = 0;
    if (a[1] < a[0])
        m = 1;
    if (a[2] < a[m])
        m = 2;
    return m;
}
 
int equal(int a[3])
{
    if (a[0] == a[1] && a[1] == a[2])
        return 1;
    return 0;
}
 
int main(void)
{
    int i;
    int a[3], cycle[3]={23, 28, 33};
    int d;
    int k = 0;
    int res;
 
    while (scanf("%d%d%d%d", &a[0], &a[1], &a[2], &d) != EOF)
    {
        if (a[0] == -1 && a[1] == -1 && a[2] == -1 && d == -1)
            break;
        k ++;
 
        do {
            i = min(a);
            a[i] += cycle[i];
            if (a[i] >= d+21252)
            {
                res = 21252;
                break;
            }
            if (equal(a) && a[0]>d)
            {
                res = a[0]-d;
                break;
            }
        } while (1);
 
        printf("Case %d: the next triple peak occurs in %d days.
", k, res);
    }
 
    return 0;
}
/**************************************************************
    Problem: 1142
    User: liangrx06
    Language: C
    Result: Accepted
    Time:0 ms
    Memory:912 kb
****************************************************************/

代码2：

#include <iostream>
using namespace std;
                 
int min(int a[3])
{           
    int m = 0; 
    if (a[1] < a[0])
        m = 1;  
    if (a[2] < a[m])
        m = 2;
    return m;
}
         
int equal(int a[3]) 
{   
    if (a[0] == a[1] && a[1] == a[2])
        return 1;
    return 0;
}
 
int main(void)
{
    int i;
    int a[3], d, cycle[3]={23, 28, 33};
    int k, res;
 
    k = 0;
    while (cin >> a[0] >> a[1] >> a[2] >> d)
    {
        if (a[0] == -1 && a[1] == -1 && a[2] == -1 && d == -1)
            break;
         
        for (i=0; i<3; i++)
        {
            while (a[i] > d)
                a[i] -= cycle[i];
        }
 
        do {
            i = min(a);
            a[i] += cycle[i];
            if (a[i] >= d+21252)
            {
                res = 21252;
                break;
            }
            if (equal(a) && a[0]>d)
            {
                res = a[0]-d;
                break;
            }
        } while (1);
 
        cout << "Case "<< ++k << ": the next triple peak occurs in ";
        cout << res << " days." << endl;
    }
 
    return 0;
}
/**************************************************************
    Problem: 1142
    User: liangrx06
    Language: C++
    Result: Accepted
    Time:10 ms
    Memory:1520 kb
****************************************************************/