ZOJ 3785：What day is that day?（数论）

What day is that day?

Time Limit: 2 Seconds Memory Limit: 65536 KB

It’s Saturday today, what day is it after $1^1 + 2^2 + 3^3 + ... + N^N$ 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.

Solve

对于$1 dots N$，对于$7$取模后，可以转换成一大串$0 dots 6$的循环，所以$1^1 + 2^2 + 3^3 + ... + N^N$可以转换成：$1^1+2^2+3^3+ dots (N\%7)^N$，然后我们可以发现，这个式子的值是七个等比数列的前$x$项和的总和。每项公比为$num^7$，第一项为$n^n$$(0leq n leq 6)$
那么我们只需要统计出$0dots 6$的个数，并利用等比数列的前$n$项和公式计算即可
_{百度了一下，发现似乎写麻烦了，貌似可以直接打表找规律？而且代码还特别短，想看短代码的自行百度吧，貌似没有找到我这种方法写的QAQ}

Code

/*************************************************************************

	 > Author: WZY
	 > School: HPU
	 > Created Time:   2019-04-20 09:49:05
	 
************************************************************************/
#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=(1<<30);
const ll INF=(1LL*1<<60);
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
int exgcd(int a,int b,int &x,int &y)
{
	if(!b)
	{
		x=1;y=0;
		return a;
	}
	else
	{
		int r=exgcd(b,a%b,y,x);
		y-=x*(a/b);
		return r;
	}
}
int inv(int a,int n)
{
	int x,y;
	exgcd(a,n,x,y);
	x=(x%n+n)%n;
	return x;
}
int Pow(int a,int b,int c)
{
	int res=1;
	while(b)
	{
		if(b&1)
			res=res*a%c;
		a=a*a%c;
		b>>=1;
	}
	return res;
}
int a[10];
int sum[10];
int main(int argc, char const *argv[])
{
	#ifndef ONLINE_JUDGE
	    freopen("in.txt", "r", stdin);
	    freopen("out.txt", "w", stdout);
	#endif
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t;
	// 计算第一项的值
	for(int i=1;i<7;i++)
		sum[i]=Pow(i,7,7);
	cin>>t;
	int n;
	while(t--)
	{
		ms(a,0);
		cin>>n;
		// 统计0~6出现的个数
		for(register int i=1;i<7;i++)
			a[i]=n/7+(n%7>=i);
		// 0忽略，初始值为1的个数
		int ans=a[1];
		int __=0;
		for(register int i=2;i<7;i++)
		{
			// 利用前n项和公式+逆元计算对7取模的结果
			int _=(Pow(i,i,7)*((Pow(sum[i],a[i],7)-1+7)%7)*inv(sum[i]-1,7)+7)%7;
			__+=_;
			if(a[i]==0)
				break;
		}
		ans=(ans+__)%7;
		int cnt=(6+ans)%7;
		if(cnt==0)
			cout<<"Sunday
";
		if(cnt==1)
			cout<<"Monday
";
		if(cnt==2)
			cout<<"Tuesday
";
		if(cnt==3)
			cout<<"Wednesday
";
		if(cnt==4)
			cout<<"Thursday
";
		if(cnt==5)
			cout<<"Friday
";
		if(cnt==6)
			cout<<"Saturday
";
	}
	return 0;
}