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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.

題意：

今天星期六，求1^1+2^2……N^N天后是星期幾

思路：

同余與模算術(shù)，利用快速冪取模的算法，時間復(fù)雜度為O(logn)。

1.先用快速冪求出1¹ ， 2² +，3³ ， ... ，N^N??????

^{對7取模之后的結(jié)果，發(fā)現(xiàn)循環(huán)節(jié)長度為}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為一組 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]區(qū)間每個數(shù)n的(n^n)%7,再求a數(shù)組的前綴和b數(shù)組，

sum表示一個循環(huán)節(jié)所貢獻(xiàn)的天數(shù)，即sum=（1^1+2^2+3^3+......+41^41+41^42）%7=6;

對于每一個樣例n，直接計算即可

AC代碼：

#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)//快速冪取模 {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++)//循環(huán)節(jié)為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;//一個循環(huán)節(jié)貢獻(xiàn)的天數(shù)int num=n/42;//循環(huán)節(jié)的個數(shù)int ans=(sum*num%7+b[n%42]%7)%7;printf("%s\n",s[ans]);}return 0; } 另一種計算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)//快速冪取模 {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++)//循環(huán)節(jié)為42{a[i]=fast_pow(i,i,7);b[i]=b[i-1]+a[i];sum=(sum%7+a[i]%7)%7;//重點理解}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; }

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