將下列ieee短浮點數轉換為十進制數，UVA495 Fibonacci Freeze【大數+萬進制】

The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...) are defined by the recurrence:

????? F0 = 0

將下列ieee短浮點數轉換為十進制數、????? F1 = 1

????? Fi = Fi?1 + Fi?2 for all i ≥ 2

? Write a program to calculate the Fibonacci Numbers.

9999二進制有多少個1？Input

The input to your program would be a sequence of numbers smaller or equal than 5000, each on a separate line, specifying which Fibonacci number to calculate.

Output

十六進制和二進制哪個大？Your program should output the Fibonacci number for each input value, one per line.

Sample Input

5

哪個進制數最大。7

11

Sample Output

十進制大還是二進制大。The Fibonacci number for 5 is 5

The Fibonacci number for 7 is 13

The Fibonacci number for 11 is 89

問題鏈接：UVA495 Fibonacci Freeze

問題簡述：（略）

問題分析：

? 大數計算問題。

程序說明：

? 用二維數組來表示大數，并且采用萬進制，高位在每一行的右邊（低位在左邊）。

? 其他都是套路。

題記：（略）

參考鏈接：（略）

AC的C++語言程序如下：

/* UVA495 Fibonacci Freeze */#include <bits/stdc++.h>using namespace std;const int N = 5000;
const int N2 = 300;
const int BASE = 1e4;
int fib[N+ 1][N2 + 1];void maketable()
{memset(fib, 0, sizeof(fib));fib[0][0] = 0;fib[1][0] = 1;for ( int i = 2 ; i <= N ; i++ ) {for ( int j = 0 ; j <= N2 ; j++ )fib[i][j] = fib[i-2][j] + fib[i-1][j];for ( int j = 0 ; j <= N2 ; j++ ) {fib[i][j+1] += fib[i][j] / BASE;fib[i][j] %= BASE;}}
}int main()
{maketable();int n;while(~scanf("%d", &n)) {int i = N2;while ( i >= 1 && !fib[n][i] )i --;printf("The Fibonacci number for %d is ", n);printf("%d", fib[n][i--]);while ( i >= 0 )printf("%04d", fib[n][i--]);printf("\n");}return 0;
}