快速幂 HDU-1021 Fibonacci Again , HDU-1061 Rightmost Digit , HDU-2674 N！Again

Posted 2021-01-26 czc1999

1.   Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 77116    Accepted Submission(s): 34896

Problem Description

There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

  

Input

Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

 

Output

Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

 

Sample Input

0

1

2

3

4

5

 

Sample Output

no

no

yes

no

no

no

数据级别在1，000，000，就正常做不会超时，最简单的快速幂，根据 (a * b) % p = (a % p * b % p) % p 。

#include <iostream>
#include <stdio.h>
using namespace std;

int main(){
	int n;
	while(scanf("%d",&n)!=EOF){
		int f0=7,f1=11;
		int fn;
		for(int i=2;i<=n;i++){
			fn=(f0+f1)%3;
			f0=f1%3;
			f1=fn%3;
		}
		if(n<=1){
			cout<<"no"<<endl;
		}
		else{
			if(fn%3==0)
			cout<<"yes"<<endl;
			else
			cout<<"no"<<endl;
		}
	}
	return 0;
}

　　

2.  Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 71510    Accepted Submission(s): 26554

 

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

 

Output

For each test case, you should output the rightmost digit of N^N.

 

Sample Input

2

3

4

 

Sample Output

7

6

---

 

 数据级别10亿，上一种就无法完成了，需要用另一种方法降低n的规模。

#include <iostream>
#include <stdio.h>
using namespace std;

int main(){
	int k;
	cin>>k;
	while(k--){
	long long n;
		cin>>n;
		long long res=1;long long a=n,b=n;
		while(b){
			if(b&1)
				res=(res*a)%10;
			a=(a*a)%10;
			b/=2;
		}
		cout<<res<<endl;
	}	
	return 0;
}

---

 

3.  N！Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6338    Accepted Submission(s): 3325

Problem Description

WhereIsHeroFrom:             Zty, what are you doing ?
Zty:                                     I want to calculate N!......
WhereIsHeroFrom:             So easy! How big N is ?
Zty:                                    1 <=N <=1000000000000000000000000000000000000000000000…
WhereIsHeroFrom:             Oh! You must be crazy! Are you Fa Shao?
Zty:                                     No. I haven‘s finished my saying. I just said I want to calculate N! mod 2009


Hint : 0! = 1, N! = N*(N-1)!

Input

Each line will contain one integer N(0 <= N<=10^9). Process to end of file.

 

Output

For each case, output N! mod 2009

 

Sample Input

4

5

 

Sample Output

24

120

---

 

 10^9这种数据级别已经不是正常做法能解出来的了，一定有特殊数据，试一下就能找到。

#include <iostream>
#include <stdio.h>
using namespace std;

int main() {
	long long n;
	while(scanf("%lld",&n)!=EOF){
		long long res=1;
		if(n>=41)
		cout<<0<<endl;
		else{
		　　for(int i=1;i<=n;i++){
			res=(res*i)%2009;
		}
		cout<<res<<endl;
		}
	}
	return 0;
}

 各种快速幂很详细：https://yq.aliyun.com/wenji/254837