ZOJ3785 What day is that day? 快速幂+找规律

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.

题意：

今天星期六，求1^1+2^2……N^N天后是星期几

思路：

同余与模算术，利用快速幂取模的算法，时间复杂度为O(logn)。

1.先用快速幂求出1¹ ， 2² +，3³ ， ... ，N^N

^{对7取模之后的结果，发现循环节长度为}42，即 (1^1)%7=(43^43)%7, (2^2)%7=(44^44)%7, (3^3)%7=(45^45)%7,

(n^n)%7=( (42+n)^(42+n) )%7

#include<bits/stdc++.h>
using namespace std;
int fast_pow(int a,int n,int mod)
{
    int ret=1;
    while(n)
    {
        if(n&1)
            ret=ret*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return ret;
}
int main()
{
    for(int i=1;i<=1000;i++)
    {
        printf("%d,",fast_pow(i,i,7));
        if(i%7==0)
            printf("\n");
    }
    return 0;
}
//以42为一组

1,4,6,4,3,1,0,
1,1,4,2,1,6,0,
1,2,5,1,5,1,0,
1,4,1,4,4,6,0,
1,1,3,2,6,1,0,
1,2,2,1,2,6,0,

1,4,6,4,3,1,0,
1,1,4,2,1,6,0,
1,2,5,1,5,1,0,
1,4,1,4,4,6,0,
1,1,3,2,6,1,0,
1,2,2,1,2,6,0,

1,4,6,4,3,1,0,
1,1,4,2,1,6,0,
1,2,5,1,5,1,0,
1,4,1,4,4,6,0,
1,1,3,2,6,1,0,
1,2,2,1,2,6,0,

1,4,6,4,3,1,0,
1,1,4,2,1,6,0,
1,2,5,1,5,1,0,
1,4,1,4,4,6,0,
1,1,3,2,6,1,0,
1,2,2,1,2,6,0,

1,4,6,4,3,1,0,
1,1,4,2,1,6,0,
1,2,5,1,5,1,0,
1,4,1,4,4,6,0,
1,1,3,2,6,1,0,
1,2,2,1,2,6,0,

1,4,6,4,3,1,0,
1,1,4,2,1,6,0,
1,2,5,1,5,1,0,
1,4,1,4,4,6,0,
1,1,3,2,6,1,0,
1,2,2,1,2,6,0,

1,4,6,4,3,1,0,
1,1,4,2,1,6,0,
1,2,5,1,5,1,0,
1,4,1,4,4,6,0,
1,1,3,2,6,1,0,
1,2,2,1,2,6,0,

1,4,6,4,3,1,0,
1,1,4,2,1,6,0,
1,2,5,1,5,1,0,
1,4,1,4,4,6,0,
1,1,3,2,6,1,0,
1,2,2,1,2,6,0,

2.然后打表求出[1,42]区间每个数n的(n^n)%7,再求a数组的前缀和b数组，

sum表示一个循环节所贡献的天数，即sum=（1^1+2^2+3^3+......+41^41+41^42）%7=6;

对于每一个样例n，直接计算即可

AC代码：

#include<bits/stdc++.h>
using namespace std;
char s[10][10]={"Saturday","Sunday","Monday","Tuesday","Wednesday","Thursday","Friday"};
int fast_pow(int a,int n,int mod)//快速幂取模
{
    int ret=1;
    while(n){
        if(n&1)
            ret=ret*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return ret;
}
int a[50];//a[n]表示(n^n)%7
int b[50];//b[n]表示a[1]+a[2]+....+a[n]
int main()
{
	ios::sync_with_stdio(0);
	b[0]=0;

    for(int i=1;i<=42;i++)//循环节为42
    {
        a[i]=fast_pow(i,i,7);
        b[i]=b[i-1]+a[i];
    }
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        int sum=b[42]%7;//一个循环节贡献的天数
        int num=n/42;//循环节的个数
        int ans=(sum*num%7+b[n%42]%7)%7;
         printf("%s\n",s[ans]);
    }
	return 0;
}

另一种计算sum的方法：

#include<bits/stdc++.h>
using namespace std;
char s[10][10]={"Saturday","Sunday","Monday","Tuesday","Wednesday","Thursday","Friday"};
int fast_pow(int a,int n,int mod)//快速幂取模
{
    int ret=1;
    while(n){
        if(n&1)
            ret=ret*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return ret;
}
int a[50];//a[n]表示(n^n)%7
int b[50];//b[n]表示a[1]+a[2]+....+a[n]
int main()
{
	ios::sync_with_stdio(0);
	b[0]=0;
	int sum=0;
    for(int i=1;i<=42;i++)//循环节为42
    {
        a[i]=fast_pow(i,i,7);
        b[i]=b[i-1]+a[i];
        sum=(sum%7+a[i]%7)%7;//重点理解
    }
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        int ans=(sum*(n/42)%7+b[n%42]%7)%7;
         printf("%s\n",s[ans]);
    }
	return 0;
}