poj 1006 Biorhythms（中国剩余定理水题）

た入场券 2024-04-19 15:26 118阅读 0赞

> 题目衔接：[http://poj.org/problem?id=1006][http_poj.org_problem_id_1006]

> <table> 
>  <tbody> 
>   <tr> 
>    <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
>    <td>&nbsp;</td> 
>    <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
>   </tr> 
>   <tr> 
>    <td><strong>Total Submissions:</strong>&nbsp;152946</td> 
>    <td>&nbsp;</td> 
>    <td><strong>Accepted:</strong>&nbsp;49676</td> 
>   </tr> 
>  </tbody> 
> </table>
> 
> 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][]

> 题目大意：每个人都有身体，情感，智力的高峰期，现在给你一个人的三个周期峰值，再给你一个天数  
> 问你下一个峰值在同一天距离给你的天数的多少天后  
> 思路:假设在x天后三个峰期有相同值给你p e i则可以得到同余方程组
> 
> x=23\*t1+p
> 
> x=28\*t2+e
> 
> x=33\*t3+i
> 
> t为常数
> 
> 然后解一下即可

代码：

/*
    题目大意：给你一个人的三个周期峰值，再给你一个天数
    问你下一个峰值全在一天距离给你的天数的多少天后
    思路:给你p e i则可以得到方程组
    下次出现峰值的方程式：
    d1=p+23*x1
    d1=e+28*x2
    d1=i+33*x3
    */
    #include<set>
    #include<map>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<cmath>
    #include<string>
    #include<vector>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    typedef long long ll;
    #define inf 0x3f3f3f
    #define bug  printf("bug\n")
    const int maxn=1e4+10;
    const double pi=acos(-1.0);
    const double esp=1e-6;
    const int N=2e2+10;
    ll m[maxn],r[maxn];
    int n;
    ll ex_gcd(ll a,ll b,ll &x,ll &y)
    {
        if(b==0)
        {
            x=1;
            y=0;
            return a;
        }
        ll d=ex_gcd(b,a%b,x,y);
        ll temp=x;
        x=y;
        y=temp-a/b*y;
        return d;
    }
    ll crt()
    {
        ll a,b,a1,b1,gcd,d,x,y,d1;
        a=m[1];
        b=r[1];
        for(int i=2; i<=3; i++)
        {
            a1=m[i];
            b1=r[i];
            d=b1-b;
            gcd=ex_gcd(a,a1,x,y);
            if(d%gcd)
                return -1;
            d1=a1/gcd;
            x=((x*d/gcd)%d1+d1)%d1;
            b+=a*x;
            a*=d1;
        }
        return b;
    
    }
    
    int main()
    {
        int p,e,i,d,q=1;
        while(scanf("%d%d%d%d",&p,&e,&i,&d)!=EOF)
        {
            if(p==-1&&e==-1&&i==-1&&d==-1)
                break;
            m[1]=23,m[2]=28,m[3]=33;
            r[1]=p,r[2]=e,r[3]=i;
            ll ans=crt();
           ans-=d;
            while(ans<=0)
                ans+=21252;
            ans%=21252;
            if(ans==0)
                ans=21252;
            printf("Case %d: the next triple peak occurs in %d days.\n",q++,ans);
        }
        return 0;
    }

[http_poj.org_problem_id_1006]: http://poj.org/problem?id=1006
[East Central North America 1999]: http://poj.org/searchproblem?field=source&key=East+Central+North+America+1999