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題目鏈接;

http://acm.hdu.edu.cn/showproblem.php?pid=1021

題目類型：

斐波那契數列

題意描述：

一個斐波那契數列，如果對3取余為0，輸出yes，反之輸出no。

解題思路：

先在全局變量定一個較大值Max為1000010，然後打表，然後在寫多實例輸入，判斷輸出即可。

題目：

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60963 Accepted Submission(s): 28487

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 <stdio.h>
# define MAX 1000010

int a[MAX];

int main ()
{
    int i,n;
    a[0]=7%3; a[1]=11%3;
    for(i=2;i<MAX;i++)
        a[i]=(a[i-1]+a[i-2])%3;
    while(scanf("%d",&n)!=EOF)
    {
        
 
if(a[n]==0)
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;
}

hdu-1021 Fibonacci Again