HDU 1021 Fibonacci Again
2015-03-10 16:59
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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
Sample Output
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 题意:对于输入的数字。计算f 是否能整除3;这道题的数据如果按照正常的方法计算,会特别大,所以需要一边计算一边处理数据。思路:如果f[i]%3==0则在下一次的f[i+1]=f[i-1]+f[i]中,f[i]可以变为0;just soso。。。
#include<stdio.h> #define N 1000000 int f[N]; int main() { int n,i; f[0]=7; f[1]=11; while(scanf("%d",&n)!=EOF) { for(i=2;i<=n;i++) { f[i]=f[i-1]+f[i-2]; if(f[i]%3==0) f[i]=0; else f[i]=f[i]%3; } if(f[n]%3==0) { printf("yes\n"); } else { printf("no\n"); } } return 0; }
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