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;}
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;}
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
