1 #include<bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 ll m=23*28*33,a[4],b,n;
 5 int mm[4];
 6 int read()
 7 {
 8     int x=0,f=1;
 9     char ch=getchar();
10     while(ch<'0'||ch>'9')
11     {
12         if(ch=='-') f=-1;
13         ch=getchar();
14     }
15     while(ch>='0'&&ch<='9')
16     {
17         x=x*10+ch-'0';
18         ch=getchar();
19     }
20     return x*f;
21 }
22 void write(int x)
23 {
24     if(x<0)
25     {
26         putchar('-');
27         x=-x;
28     }
29     if(x>9) write(x/10);
30     putchar(x%10+'0');
31 }
32 void exgcd(ll a,ll b,ll &d,ll &x,ll &y)
33 {
34     if(b==0)
35     {
36         d=a;
37         x=1;
38         y=0;
39     }
40     else
41     {
42         exgcd(b,a%b,d,x,y);
43         int t=x;
44         x=y;
45         y=t-a/b*y;
46     }
47 }
48 int main()
49 {
50     mm[1]=23;mm[2]=28;mm[3]=33;int k=0;
51     while(scanf("%lld%lld%lld%lld",&a[1],&a[2],&a[3],&b)!=EOF)
52     {
53         k++;ll ans=0;
54         if(a[1]==-1)break;
55         for(int i=1;i<=3;i++)
56         {
57             ll d,x,y,mi=m/mm[i];
58             exgcd(mi,mm[i],d,x,y);
59             ans=(ans+mi*x*a[i])%m;
60         }
61         ll xx=(ans+m)%m;
62         if(xx>b)n=xx-b;
63         else if(ans<=b)
64         {
65             for(int i=1;;i++)
66             {
67                 if(ans+i*m>b)
68                 {
69                     n=ans+i*m-b;
70                     break;
71                 }
72             }
73         }
74         else
75         {
76             for(int i=-1;;i--)
77             {
78                 if(ans+i*m<b)
79                 {
80                     n=ans+(i+1)*m-b;
81                     break;
82                 }
83             }
84         }
85         printf("Case %d: the next triple peak occurs in %lld days.\n",k,n);
86     }
87     return 0;
88 }

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const ll lcm=21252;//21252=23*28*33
 5 ll m[4],a[4],d;
 6 void exgcd(ll a,ll b,ll &gcd,ll &x,ll &y)
 7 {
 8     if(!b)
 9     {
10         gcd=a;
11         x=1;
12         y=0;
13     }
14     else
15     {
16         exgcd(b,a%b,gcd,x,y);
17         ll t=x;x=y;y=t-a/b*y;
18     }
19 }
20 int main()
21 {
22     int cnt=0;
23     m[1]=23;m[2]=28;m[3]=33;
24     while(~scanf("%lld%lld%lld%lld",&a[1],&a[2],&a[3],&d))
25     {
26         cnt++;
27         ll ans=0,Mi,gcd,x,y;
28         if(a[1]==0&&a[2]==0&&a[3]==0&&d==0)//数据最大不超过21252 ，所以可以直接输出
29         {
30             printf("Case %d: the next triple peak occurs in %lld days.\n",cnt,lcm);
31             continue;
32         }
33         if(a[1]==-1)//写一个就够了，没必要全写出来，因为题目已说明所有给定时间是非负的
34         break;
35         for(int i=1;i<=3;i++)//中国剩余定理的应用
36         {
37             Mi=lcm/m[i];
38             exgcd(Mi,m[i],gcd,x,y);
39             ans=((ans+Mi*x*a[i])%lcm+lcm)%lcm;
40         }
41         ans=((ans%lcm)-(d%lcm)+lcm)%lcm;
42         if(ans<=0) ans+=lcm;//若小于0，直接加上21252，还是因为数据最大不超过21252
43         printf("Case %d: the next triple peak occurs in %lld days.\n",cnt,ans);
44     }
45     return 0;
46 }