poj1006 Biorhythms

最新推荐文章于 2019-08-07 08:00:00 发布

原创最新推荐文章于 2019-08-07 08:00:00 发布 · 198 阅读

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中国剩余定理

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本文介绍了一个基于中国剩余定理的问题解决方法，通过给定个人生命中三个周期（体力、情绪和智力）的峰值日期，计算下一个这三个周期同时达到峰值的日子。文章提供了详细的数学公式和C语言实现代码。

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.

Solution

题目巨长，题意巨水

中国剩余定理

给出一堆形如

x \equiv a 1 (mod m 1)

$x\equiv a_1\pmod {m_1}$

x \equiv a 2 (mod m 2)

$x\equiv a_2\pmod {m_2}$

x \equiv a 3 (mod m 3)

$x\equiv a_3\pmod {m_3}$
这样的方程，求一个最小的x

定义 $M=\prod_{i=1}^{n}{m_i}$ ，且 $M_i=\frac{M}{m_i}$ ，有

a n s = (\sum i = 1 n M i * M - 1 i * a i) (mod M)

$ans=(\sum_{i=1}^{n}{M_i*M_i^{-1}*a_i})\pmod M$
其中

M−1iMi−1 $M_i^{-1}$ 表示

MiMi $M_i$ 在模

mimi $m_i$ 意义下的逆元，这里

mimi $m_i$ 不一定是质数因此要exgcd

Code

#include <stdio.h>
#include <string.h>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)

typedef long long LL;
const int N=2005;

LL m[N],a[N];

LL ksm(LL x,LL dep,LL MOD) {
    if (dep==0) return 1;
    if (dep==1) return x;
    LL tmp=ksm(x,dep/2,MOD);
    if (dep&1) return tmp*tmp%MOD*x%MOD;
    return tmp*tmp%MOD;
}

void exgcd(LL a,LL b,LL &x,LL &y)  {
    if (!b) {
        x=1; y=0;
        return ;
    }
    exgcd(b,a%b,x,y);
    LL tmp=x; x=y; y=tmp-(a/b)*y;
}

int main(void) {
    int cnt=0;
    while (1) {
        LL M=1,ans=0,st;
        rep(i,1,3) scanf("%lld",&a[i]);
        scanf("%lld",&st);
        if (a[1]==-1&&a[2]==-1&&a[3]==-1&&st==-1) break;
        m[1]=23; m[2]=28; m[3]=33;
        rep(i,1,3) M=M*m[i];
        rep(i,1,3) {
            LL Mi=M/m[i];
            LL x,y; exgcd(Mi,m[i],x,y);
            ans=(ans+Mi%M*x%M*a[i]%M+M)%M;
        }
        ans-=st;
        if (ans<=0) ans=ans+M;
        printf("Case %d: the next triple peak occurs in %lld days.\n",++cnt,ans);
    }
    return 0;
}