J-what Day Is this day? (Watch for rules)

It ' s Saturday today, what the it after + + + ... + NN days? Input

There are multiple test cases. The "a" 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.

The general meaning of the topic is as follows:

is to give you one + + + + ... + nn this formula, and tell you now is Saturday, want you to judge in one + + + + ... + nn after so many days is the day of the week.

At first I thought that if I wrote directly, I would have timed out. Then that it must have a regular, then the first I was to calculate the value by hand, but after 7 to find that it is impossible to continue, and then I know that you can use the table to find the cycle of this method to find the law, so go to the system to learn a bit.

#include <stdio.h>
#include <string.h>
int main () {
	char word[100];
	scanf ("%s", word);
	int Len=strlen (word);
	I represents the cycle for 
	(int i=1;i<=len;i++) if (len%i==0) {	//here because I'm asking for a string cycle, so if I can be divisible by Len, then it's a cycle, and then the following judgment ; 
		//ok used to mark; 
		int ok=1;
		for (int j=i;j<len;j++)
		//The following statement is the key. Determine if the first j is equal to the previous one; 
			if (word[j]!=word[j%i])   {ok=0; break;}}
	}

So the above method can be used to find the problem of the cycle is 294, so good to do;

#include <stdio.h>
#include <string.h>
#include <string>
using namespace std;
Char ss[8][10]={"Saturday", "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday"};
int work (int n) {
	int sum=1;
	for (int i=1;i<=n;i++) {
		sum=sum*n;
		sum=sum%7;
	}
	return sum;
}
int main () {
	int t,n;
	int f[301]={0};
	for (int i=1;i<=294;i++) {
		f[i]=f[i-1]+work (i);
		f[i]=f[i]%7;
	}
	scanf ("%d", &t);
	while (t--) {
		scanf ("%d", &n);
		n=n%294;
		printf ("%d\n", F[n]);
		int t=f[n];
		printf ("%s\n", Ss[t]);
	}

Very useful way, haha.