ACM刷题之ZOJ————What day is that day?

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.






初看这题目，以为是快速幂，但看了范围后，感觉应该是规律题。
总结下做规律题的方法，尽可能的多大几组数据（要保证数据是正确的）
然后放到sublime里面找字串就好了。这题我找出的规律是294 但是网上说规律是40多
可能找的太急，找到了规律的倍数吧


下面是ac代码（里面有快速幂取模的代码）
#include<stdio.h>

int qmod(int a,int b)
{
	int r=1;
	while(b)
	{
		if(b&1)
			r=((r%7)*(a%7))%7;
		a=((a%7)*(a%7))%7;
		b>>=1;
		
	}
	return r;
}
int x[400];
int main()
{
	int zu,c=0,n,q;
	long long k;
	for(int i=1;i<300;i++)
	{
		//if(!(i%7))
		//PRintf("\n\n");
		c+=qmod(i,i);
		x[i]=(c%7);
	}
	scanf("%d",&n);
	while(n--)
	{
		scanf("%lld",&k);
		k=k%294;
		if(x[k]==0)
		{
			printf("Saturday\n");
		}else if(x[k]==1)
		{
			printf("Sunday\n");
		}else if(x[k]==2)
		{
			printf("Monday\n");
		}else if(x[k]==3)
		{
			printf("Tuesday\n");
		}else if(x[k]==4)
		{
			printf("Wednesday\n");
		}else if(x[k]==5)
		{
			printf("Thursday\n");
		}else if(x[k]==6)
		{
			printf("Friday\n");
		}
		
		
	}
	
	
}