ZOJ 3785 What day is that day?（数论：费马小定理）

What day is that day?

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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It's Saturday today, what day is it after 11 + 22 + 33 + ... + NN 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.

分析：运用费马小定理，简化运算并求出循环节

1^1     2^2     3^3     4^4     5^5     6^6     7^7
8^8     9^9     10^10   11^11   12^12   13^13   14^14
15^15   16^16   17^17   18^18   19^19   20^20   21^21
22^22   23^23   24^24   25^25   26^26   27^27   28^28
29^29   30^30   31^31   32^32   33^33   34^34   35^35
36^36   37^37   38^38   39^39   40^40   41^41   42^42
43^43   44^44   45^45   46^46   47^47   48^48   49^49
转化后-->>
1^1     2^2     3^3      4^4     5^5     6^6      0
1^2     2^3     3^4      4^5     5^6     6^1      0 
1^3     2^4     3^5      4^6     5^1     6^2      0
1^4     2^5     3^6      4^1     5^2     6^3      0
1^5     2^6     3^1      4^2     5^3     6^4      0
1^6     2^1     3^2      4^3     5^4     6^5      0
1^1     2^2     3^3      4^4     5^5     6^6      0

#include<iostream>
#include<math.h>
#define ll long long
using namespace std;
int s[50];
void chose(ll x){
        switch(x){
                case 0:printf("Sunday
");break;
                case 1:printf("Monday
");break;
                case 2:printf("Tuesday
");break;
                case 3:printf("Wednesday
");break;
                case 4:printf("Thursday
");break;
                case 5:printf("Friday
");break;
                case 6:printf("Saturday
");break;
        }
}
int main()
{
        int T;
        ll n;
        for(int i=1;i<=42;i++){
                int a,p,sum=1;
                p=(i%6)?i%6:6;
                //为6的倍数情况下保留一个六
                a=i%7;
                for(int j=1;j<=p;j++)
                        sum=(sum*a)%7;
                s[i]=(s[i-1]+sum)%7;
        }
        scanf("%d", &T);
        while(T--){
                scanf("%lld", &n);
                ll ans=(n/42%7*s[42]%7+s[n%42])%7;
                ans=(ans+6)%7;
                chose(ans);
        }
        return 0;
}