What day is that day?

转：https://blog.csdn.net/menxiaoyuan/article/details/50451719 了解了如何去找循环节 What day is that day? 题解：

Description

It’s Saturday today, what day is it after 1¹ + 2² + 3³ + … + N^N days?

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There is only one line containing one integer N (1 <= N <= 1000000000).

Output

For each test case, output one string indicating the day of week.

Sample Input

2
1
2

Sample Output

Sunday
Thursday

Hint

A week consists of Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

[cpp] view plain copy

print ?

1. //打表代码–找循环节   
2. #include <iostream>  
3. #include <cstdio>  
4. #include <cstring>  
5. #include <algorithm>  
6. using namespace std;  
7. const int maxn = 11000;  
8. int a[maxn], b[maxn], f[maxn];   
9. int main(){  
10.      int t, i, j, n;  
11.      int sum, x, cnt, ad;  
12.      while(cin>>t){  
13.          while(t–){  
14.              //cin>>n;  
15.              sum = 0;  
16.              memset(a, 0, sizeof(a));  
17.              memset(b, 0, sizeof(b));  
18.              memset(f, 0, sizeof(f));  
19.              for(n = 1; n <= 10005; n++){  
20.                  x = 1;  
21.                  for(i = 1; i <= n; i++){  
22.                      x = x*n;  
23.                      x = x%7;  //因为数会非常大，所以利用公式，取余  
24.                  }  
25.                   sum += x;  
26.                   sum = sum%7;  //因为数会非常大，所以利用公式，取余  
27.                   a[n] = sum;   //把所有结果存入a数组  
28.              }  
29.              cnt = 1;  
30.              for(i = 1; i <= 10000; i++){  
31.                  if(a[i]==1)  
32.                      {  
33.                          b[cnt++] = i;  //把结果为1的序号 存入b  
34.                         f[i] = 1;  
35.                     }  
36.                     /*cout<<a[i]; 
37.                     if(i%294==0) 
38.                        cout<<endl<<endl;*/  
39.             }  
40.             for(i = 2; i < cnt; i++){  
41.                 ad = b[i]-1;  //每一个 的间隔  
42.                 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  
43.                 {  
44.                     if(f[b[i]+ad*j] != 1)  
45.                         break;  
46.                 }  
47.                if(j==7){  
48.                    cout<<b[i]<<endl;  
49.                     break;  
50.                 }  
51.             }  
52.         }  
53.     }  
54.     return 0;  
55. }  

//打表代码--找循环节 

}

[cpp] view plain copy

print ?

1. //ac代码   
2. #include<iostream>  
3. #include<cstdio>  
4. #include<cstring>  
5. using namespace std;  
6. int e[300];  
7. char s[10][20]={“Saturday”,“Sunday”, “Monday”, “Tuesday”, “Wednesday”, “Thursday”, “Friday”};  
8. int main(){  
9.     int T,n,i,j,x,sum=0;  
10.     for(i=1;i<=294;i++){  
11.         x=1;  
12.         for(j=1;j<=i;j++){  
13.             x*=i;  
14.             x%=7;  
15.         }  
16.         sum+=x;  
17.         sum%=7;  
18.         e[i]=sum;  
19.     }  
20.     cin >> T;  
21.     while(T–){  
22.         cin >> n;  
23.         n%=294;  
24.         if(n==0) n=294;  
25.         cout << s[e[n]] << endl;  
26.     }  
27.     return 0;  
28. }  

pre>

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Description

It’s Saturday today, what day is it after 1¹ + 2² + 3³ + … + N^N days?

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There is only one line containing one integer N (1 <= N <= 1000000000).

Output

For each test case, output one string indicating the day of week.

Sample Input

2
1
2

Sample Output

Sunday
Thursday

Hint

A week consists of Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

[cpp] view plain copy

print ?

1. //打表代码–找循环节   
2. #include <iostream>  
3. #include <cstdio>  
4. #include <cstring>  
5. #include <algorithm>  
6. using namespace std;  
7. const int maxn = 11000;  
8. int a[maxn], b[maxn], f[maxn];   
9. int main(){  
10.      int t, i, j, n;  
11.      int sum, x, cnt, ad;  
12.      while(cin>>t){  
13.          while(t–){  
14.              //cin>>n;  
15.              sum = 0;  
16.              memset(a, 0, sizeof(a));  
17.              memset(b, 0, sizeof(b));  
18.              memset(f, 0, sizeof(f));  
19.              for(n = 1; n <= 10005; n++){  
20.                  x = 1;  
21.                  for(i = 1; i <= n; i++){  
22.                      x = x*n;  
23.                      x = x%7;  //因为数会非常大，所以利用公式，取余  
24.                  }  
25.                   sum += x;  
26.                   sum = sum%7;  //因为数会非常大，所以利用公式，取余  
27.                   a[n] = sum;   //把所有结果存入a数组  
28.              }  
29.              cnt = 1;  
30.              for(i = 1; i <= 10000; i++){  
31.                  if(a[i]==1)  
32.                      {  
33.                          b[cnt++] = i;  //把结果为1的序号 存入b  
34.                         f[i] = 1;  
35.                     }  
36.                     /*cout<<a[i]; 
37.                     if(i%294==0) 
38.                        cout<<endl<<endl;*/  
39.             }  
40.             for(i = 2; i < cnt; i++){  
41.                 ad = b[i]-1;  //每一个 的间隔  
42.                 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  
43.                 {  
44.                     if(f[b[i]+ad*j] != 1)  
45.                         break;  
46.                 }  
47.                if(j==7){  
48.                    cout<<b[i]<<endl;  
49.                     break;  
50.                 }  
51.             }  
52.         }  
53.     }  
54.     return 0;  
55. }  

//打表代码--找循环节 

}

[cpp] view plain copy

print ?

1. //ac代码   
2. #include<iostream>  
3. #include<cstdio>  
4. #include<cstring>  
5. using namespace std;  
6. int e[300];  
7. char s[10][20]={“Saturday”,“Sunday”, “Monday”, “Tuesday”, “Wednesday”, “Thursday”, “Friday”};  
8. int main(){  
9.     int T,n,i,j,x,sum=0;  
10.     for(i=1;i<=294;i++){  
11.         x=1;  
12.         for(j=1;j<=i;j++){  
13.             x*=i;  
14.             x%=7;  
15.         }  
16.         sum+=x;  
17.         sum%=7;  
18.         e[i]=sum;  
19.     }  
20.     cin >> T;  
21.     while(T–){  
22.         cin >> n;  
23.         n%=294;  
24.         if(n==0) n=294;  
25.         cout << s[e[n]] << endl;  
26.     }  
27.     return 0;  
28. }  

pre>

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