UVA 10334 - Ray Through Glasses(高精度斐波那契)

Ray Through Glasses 

Suppose we put two panes of glass back-to-back. How many ways  are there for light rays to pass through or be reflected after changing direction n times ? Following figure shows the situations when the value of nis 0, 1 and 2. 
                                  

Input 

It is a set of lines with an integer n where 0 <= n <= 1000 in each of them.

Output 

For every one of these integers a line containing  as described above.

Sample Input 

012

Sample Output 

123

题意：斐波那契，前1000项。

思路：用高精度去写

代码：

#include <stdio.h>#include <string.h>int n;int f[1005][35];void solve() {memset(f, 0, sizeof(f));f[0][0] = 1; f[1][0] = 2;for (int i = 2; i <= 1000; i ++) {for (int j = 0; j < 35; j ++) {f[i][j] += f[i - 1][j] + f[i - 2][j];f[i][j + 1] += f[i][j] / 100000000;f[i][j] %= 100000000;}}}void print(int n) {int i, j;for (i = 34; i >= 0; i --) {if (f[n][i] != 0)break;}printf("%d", f[n][i]);for (j = i - 1; j >= 0; j --)printf("%08d", f[n][j]);printf("\n");}int main() {solve();while (~scanf("%d", &n)) {print(n);}return 0;}