Physiological cycle POJ 1006, physiological cycle poj1006

Physiological cycle POJ 1006, physiological cycle poj1006

Time Limit:1000 MS		Memory Limit:10000 K
Total Submissions:138101		Accepted:44225

Description

There are three physiological cycles in life: physical strength, emotional strength, and mental strength. The cycle is 23 days, 28 days, and 33 days. One day in each cycle is the peak. On the peak day, people will do well in the corresponding aspects. For example, at the peak of the intelligence cycle, people will be agile, and their energy is easy to concentrate. Because the perimeter of the three cycles is different, the peaks of the three cycles usually do not fall on the same day. For everyone, we want to know when three peaks fall on the same day. For each cycle, we will give the number of days from the first day of the current year to the peak (not necessarily the first peak time ). Your task is to specify the number of days from the first day of the current year, and the output starts from the specified time (excluding the specified time) the next three peaks fall on the same day (days from the given time ). For example, if the given time is 10 and the time of three peaks in the same day is 12 next time, 2 is output (note that this is not 3 ).

Input

Enter four integers: p, e, I, and d. P, e, And I indicate the time when the peak of physical strength, emotion, and intelligence occurred (calculated from the first day of the current year ). D is the given time, which may be less than p, e, or I. All the given time values are non-negative and less than 365, and the requested time is less than 21252.

When p = e = I = d =-1, the input data ends.

Output

The time of the next three peaks on the same day (days from the given time) from the given time ).

The format is as follows:
Case 1: the next triple peak occurs in 1234 days.

Note: even if the result is one day, the plural form is "days ".

Sample Input

0 0 0 00 0 0 1005 20 34 3254 5 6 7283 102 23 320203 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

Translator

The Program Design Practice of Peking University is 2007. We can start with the cycle length, and it is easy to see that the three cycle lengths are of mutual quality. Therefore, we can consider the Chinese Remainder Theorem, we need to construct a solution. Ans % 23 = pans % 28 = eans % 33 = I where p, e, I is the input data and then run a bare Chinese Remainder Theorem.

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<queue>using namespace std;const int MAXN=1001;inline void read(int &n){char c='+';int x=0;bool flag=0;while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}while(c>='0'&&c<='9'){x=x*10+(c-48);c=getchar();}flag==1?n=-x:n=x;}int getgcd(int a,int b){return b==0?a:getgcd(b,a%b); }int exgcd(int a,int b,int &x,int &y){if(b==0){x=1;y=0;return a;}int r=exgcd(b,a%b,x,y);int tmp=x;x=y;y=(tmp-(a/b)*y);return r;}int crt(int *a,int *m,int num){int M=1,ans=0;for(int i=1;i<=num;i++)M*=m[i];int x,y;for(int i=1;i<=num;i++){int nowm=(M/m[i]);int remain=exgcd(nowm,m[i],x,y);ans=(ans+a[i]*nowm*x)%M;}return ans;}int tl,qg,zs,day;int now=1;int a[MAXN];int m[MAXN];int main(){ios::sync_with_stdio(0);m[1]=23;m[2]=28;m[3]=33;int tot=0;while(cin>>a[1]>>a[2]>>a[3]>>day){if(a[1]==-1)break;int out=crt(a,m,3);if(out<=day)out+=21252;printf("Case %d: the next triple peak occurs in %d days.\n",++tot,out-day);}return 0;}