NUC1100 Biorhythms【中国剩余定理】

Biorhythms

时间限制: 1000ms 内存限制: 65536KB

问题描述

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.

输入描述

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.

输出描述

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

2

0 0 0 0
0 0 0 100
5 20 34 325
-1 -1 -1 -1

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 1: the next triple peak occurs in 16994 days.
Case 2: the next triple peak occurs in 8910 days.
Case 3: the next triple peak occurs in 10789 days.

来源

East Central North America 1999

问题分析：

这个题与《POJ1006 UVA756 Biorhythms》几乎完全相同，只是输入输出格式不同，核心计算代码是一样的，代码简单改一下就可以了。

程序说明：

格式很容易出错，需要注意。

参见参考链接。

参考链接：POJ1006 UVA756 Biorhythms

题记：

程序写多了，似曾相识的也就多了。

AC的C++程序如下：

#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 n;
    scanf("%d", &n);
    for(int j=1; j<=n; j++) {
        if(j != 1)
            printf("\n");

        int ncase = 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.\n", ncase++, x-d);
        }
    }

    return 0;
}