Hdu 大菲波数 简单大数相加

题意：

求出2000以内的斐波那契数

思路：大数相加

Tips：---

Code：

View Code

 1 #include <stdio.h>
 2 #include <cstring>
 3 
 4 int f[1010][220];
 5 int main()
 6 {
 7     int i, j;
 8     int n, T;
 9     int tmp, ttmp;
10     f[1][0] = f[2][0] = 1;
11     for(i = 3; i < 1010; ++i){
12         for(j = 0; j <= 210; ++j)
13             f[i][j] = f[i-1][j] + f[i-2][j];
14         tmp = 0;
15         for(j = 0; j <= 210; ++j){
16             ttmp = f[i][j] + tmp;
17             tmp = ttmp/10;
18             f[i][j] = ttmp%10;
19         }
20     }
21 
22     while(scanf("%d", &T) != EOF)
23     while(T--)
24     {
25         scanf("%d", &n);
26         i = 210;
27         while(f[n][i] == 0) i--;
28         while(i >= 0)
29         printf("%d", f[n][i--]);
30         puts("");
31     }
32 
33     return 0;
34 }

 

 

 

转载于:https://www.cnblogs.com/Griselda/archive/2012/08/06/2625652.html