zoj 3785 What day is that day? (打表找规律)

题目

思路:比赛的时候有想过找循环节,但是,打表打错了。 后来,看着过了挺多人,就急了, 看了一下别人的时间 耗时都挺长的,就以为不是找规律,

没想到真是找规律,不过,这个题的数据可能挺大的。

AC代码:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 char s[10][20] = {"Saturday", "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday"};
 7 int a[300];
 8 
 9 int main()
10 {
11     int t, i, n;
12     int sum, x;
13     sum = 0;
14     for(n = 1; n <= 294; n++)
15     {
16         x = 1;
17         for(i = 1; i <= n; i++)
18         {
19             x = x*n;
20             x = x%7;
21         }
22         sum += x;
23         sum = sum%7;
24         a[n] = sum;
25     }
26     while(cin>>t)
27     {
28         while(t--)
29         {
30             cin>>n;
31             x = n%294;
32             if(x==0)
33                 x = 294;
34             cout<<s[a[x]]<<endl;
35         }
36     }
37     return 0;
38 }

打表代码:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn = 11000;
 7 int a[maxn], b[maxn], f[maxn];
 8 
 9 int main()
10 {
11     int t, i, j, n;
12     int sum, x, cnt, ad;
13     while(cin>>t)
14     {
15         while(t--)
16         {
17             //cin>>n;
18             sum = 0;
19             memset(a, 0, sizeof(a));
20             memset(b, 0, sizeof(b));
21             memset(f, 0, sizeof(f));
22             for(n = 1; n <= 10005; n++)
23             {
24                 x = 1;
25                 for(i = 1; i <= n; i++)
26                 {
27                     x = x*n;
28                     x = x%7;  //因为数会非常大,所以利用公式,取余
29                 }
30                 sum += x;
31                 sum = sum%7;  //因为数会非常大,所以利用公式,取余
32                 a[n] = sum;   //把所有结果存入a数组
33             }
34             cnt = 1;
35             for(i = 1; i <= 10000; i++)
36             {
37                 if(a[i]==1)
38                     {
39                         b[cnt++] = i;  //把结果为1的序号 存入b
40                         f[i] = 1;
41                     }
42                     /*cout<<a[i];
43                     if(i%294==0)
44                         cout<<endl<<endl;*/
45             }
46             for(i = 2; i < cnt; i++)
47             {
48                 ad = b[i]-1;  //每一个 的间隔
49                 for(j = 1; j <= 6; j++)  //如果照这个间隔向后推 几次,还是符合间隔的话,就找到了循环节
50                 {
51                     if(f[b[i]+ad*j] != 1)
52                         break;
53                 }
54                 if(j==7)
55                 {
56                     cout<<b[i]<<endl;
57                     break;
58                 }
59             }
60         }
61     }
62     return 0;
63 }
原文地址:https://www.cnblogs.com/bfshm/p/3664766.html