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HDU 1201 Fibonacci Again
Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37229 Accepted Submission(s): 17970
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.
Print the word "no" if not.
Sample Input
012345
Sample Output
no
no
yes
no
no
no
算法分析:
一开始用的__int64 开数组挂了,估计也会挂,n的范围可以大到1000000, 如果n的值比较大,那么f[n] 就超数据类型了。
查到了这个公式:
(m+n)%3 = (m%3+n%3)%3 ;
取完余数后再存到数组里,f[n]就不会超数据类型了。
#include <stdio.h>#include <string.h>#include <stdlib.h>int f[1000001];int main(){ int n; int i, j; f[0]=7%3 ; f[1]=11%3 ; for(i=2; i<1000000; i++) { f[i]=(f[i-1]%3+f[i-2]%3)%3; } while(scanf("%d", &n)!=EOF) { if(f[n]==0) printf("yes\n"); else printf("no\n"); } return 0;}
HDU 1201 Fibonacci Again
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