# tctony

Focus on linux,emacs,c/c++,python,algorithm...

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# Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1911    Accepted Submission(s): 905

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```
#include<iostream>
using namespace std;

int main(){

int n;

while(cin>>n){

if(n==0||n==1)
cout
<<"no"<<endl;

else {

if((n-2)%4==0)
cout
<<"yes"<<endl;

else
cout
<<"no"<<endl;
}

}

return 0;
}

0ms0k内存是怎么做出来的呢？
posted on 2007-12-04 07:40 tctony 阅读(816) 评论(3)  编辑 收藏 引用

### 评论

# re: HDU 1021 2010-04-14 23:27 蓝色书虫

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>

int main(){

int i=0;
long a[10000] ={0};
a[0] =7;
a[1] =11;

for(i=0;i<=100;i++){
//第0组，第一组数据直接输出
if(i==0||i==1){
if(a[0]%3==0)
printf("%d %d yes\n",i,i%8);
else printf("%d %d no\n",i,i%8);
continue;
}
//用a【i】存储F（i）
a[i] =a[i-1]+a[i-2];
if(a[i]%3==0)
printf("%d %d yes\n",i,i%8);
else printf("%d %d no\n",i,i%8);
}//for

getch();
return 0;
}//main

# re: HDU 1021 2010-04-15 08:24 tctony
long类型溢出了  回复  更多评论

# re: HDU 1021 2010-04-15 10:25 蓝色书虫