Hat's Fibonacci
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12633 Accepted Submission(s): 4241
Problem Description
A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input, and print that Fibonacci number.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input, and print that Fibonacci number.
Input
Each line will contain an integers. Process to end of file.
Output
For each case, output the result in a line.
Sample Input
100
Sample Output
4203968145672990846840663646 Note: No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.
Author
//题目中说答案中的结果不会超过2005位数字,而我一个int 存了8位,所以可以确定数组的第二维 //最多开个260 260*8=2080>2005就行,而1维嘛,10的2006次方大约等于2的7000多次方,所以开个8000足够; #include<stdio.h> int i,j,n; int ans[8000][255]={{0}};//先打表再查表 ans[个数][这个数的每一位的值] void fun() { for(i=1;i<5;i++) ans[i][1]=1;//注意,这里为了方便,二维都是从1开始的 for(i=5;i<8000;i++) for(j=1;j<255;j++) { ans[i][j]+=ans[i-1][j]+ans[i-2][j]+ans[i-3][j]+ans[i-4][j]; ans[i][j+1]+=ans[i][j]/100000000; ans[i][j]%=100000000; } } int main() { fun(); while(scanf("%d",&n)!=EOF) { for(i=254;i>0;i--)//从前往后找到有效值 if(ans[n][i]) break; printf("%d",ans[n][i]);//为了不输出前导0,所以这里特别输出 i--; for(i;i>0;i--)//继续使用上面循环中的i,并减1 printf("%.8d",ans[n][i]); printf("\n"); } return 0; }