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
原文地址：https://www.cnblogs.com/czc1999/p/10111573.html