# poj 1006 Biorhythms

 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 125197 Accepted: 39528

Description

Input

Output

Case 1: the next triple peak occurs in 1234 days.

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

Translator

 1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cmath>
5 #include<algorithm>
6 #include<cstring>
7 using namespace std;
8 int a[10],m[10],tot,now,M;
9 inline void exgcd(int a,int b,int &d,int &x,int &y){
10     if(b==0){
11         x=1; y=0; d=a;
12         return ;
13     }
14     exgcd(b,a%b,d,y,x); y-=x*(a/b);
15 }
16 inline int China(int r){
17     M=1;
18     int Mi,d,x0=0,y0=0,ans=0;
19     for(int i=1;i<=r;i++) M*=m[i];
20     for(int i=1;i<=r;i++){
21         Mi=M/m[i];
22         exgcd(Mi,m[i],d,x0,y0);//x0是Mi的乘法逆元
23         ans=(ans+Mi*x0*a[i])%M;
24     }
25     ans=(ans+M)%M;
26     return ans;
27 }
28 int main(){
29     m[1]=23; m[2]=28; m[3]=33;
30     while(scanf("%d%d%d%d",&a[1],&a[2],&a[3],&now)&&a[1]!=-1){
31         tot++;
32         if(a[1]==a[2]&&a[2]==a[3]&&a[3]==0){
33             printf("Case %d: the next triple peak occurs in %d days.\n",tot,21252-now);
34             continue;
35         }
36         int ANS=China(3);
37         while(ANS<=now) ANS+=M;
38         printf("Case %d: the next triple peak occurs in %d days.\n",tot,ANS-now);
39     }
40     return 0;
41 }

posted @ 2016-02-27 16:37  CXCXCXC  阅读(158)  评论(0编辑  收藏  举报