ZOJ1160 Biorhythms【中国剩余定理】

Biorhythms

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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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.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

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

1

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.

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Source: East Central North America 1999; Pacific Northwest 1999

问题链接：ZOJ1160 Biorhythms。

题意简述：（略）

问题分析：

本题可以直接用中国剩余定理来解，同余方程如下：

X≡p(mod 23)

X≡e(mod 28)

X≡i(mod 33)

其中，23、28和33是两两互素的，满足中国剩余定理的前提条件。

程序说明：

这里给出两个C语言程序，一个是使用中国剩余定理来解的，另外一个是使用试探法来解（时间上可以接受，逻辑更加简单易懂）。

这个题的原题应该是UVALive5421，然而输入输出格式不一样。

参考链接：POJ1006 UVA756 UVALive5421 Biorhythms【中国剩余定理】

AC的C语言程序如下：

/* ZOJ1160 Biorhythms */

#include <stdio.h>

// 递推法实现扩展欧几里德算法
long exgcd(long a, long b, long  *x, long *y)
{
    long x0=1, y0=0, x1=0, y1=1;
    long r, q;
    *x=0;
    *y=1;

    r = a % b;
    q = (a - r) / b;
    while(r)
    {
        *x = x0 - q * x1;
        *y = y0 - q * y1;
        x0 = x1;
        y0 = y1;
        x1 = *x;
        y1 = *y;

        a = b;
        b = r;
        r = a % b;
        q = (a - r) / b;
    }
    return b;
}

// 扩展欧几里德算法求逆元
long minv(long a, long p)
{
    long x, y;

    exgcd(a, p, &x, &y);

    return x<0 ? x+p : x;
}

// 中国剩余定理(Chinese remainder theorem, CRT)
long crt(long a[], long m[], int n)
{
    long bm=1, mi, x=0;
    int i;

    for(i=0; i<n; i++)
        bm *= m[i];
    for(i=0; i<n; i++) {
        mi = bm / m[i];
        x += a[i] * mi * minv(mi, m[i]);
        x %= bm;
    }

    return x;
}

int main(void)
{
    long p, e, i, d;
    long a[3], m[3];
    long x, bm;
    int t;

    scanf("%d", &t);
    while (t--) {
        int caseno = 1;
        for(;;) {
            scanf("%ld%ld%ld%ld", &p, &e, &i, &d);
            if(p==-1 && e==-1 && i==-1 && d==-1)
                break;
            a[0] = p;
            a[1] = e;
            a[2] = i;
            m[0] = 23;
            m[1] = 28;
            m[2] = 33;

            bm = 23 * 28 * 33;
            x = crt(a, m, 3);
            while(x<=d)
                x += bm;

            printf("Case %d: the next triple peak occurs in %ld days.
", caseno++, x-d);
        }
        if(t)
            printf("
");
    }

    return 0;
}

AC的C语言程序如下：

/* ZOJ1160 Boirhythm */

#include <stdio.h>

int main(void)
{
    int t, p, e, i, d, date;

    scanf("%d", &t);
    while(t--) {
        int caseno = 1;
        while(~scanf("%d%d%d%d", &p, &e, &i, &d))
        {
            if(p==-1)
                break;
            date=d;
            d++;
            while((d  -p) % 23 != 0)
                d++;
            while((d - e) % 28 !=0 || (d - i) % 33 != 0)
                d += 23;
            printf("Case %d: the next triple peak occurs in %d days.
", caseno++, d - date);
        }
        if(t)
            printf("
");
    }

    return 0;
}