HDU - 1865 1string（大数）

题目链接：http://acm.hust.edu.cn/vjudge/contest/121397#problem/F

http://acm.hdu.edu.cn/showproblem.php?pid=1865

Description

You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your work is to find the total number of result you can get.

Input

The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.

Output

The output contain n lines, each line output the number of result you can get .

Sample Input
3
1
11
11111
Sample Output
1
2
8

AC代码:

 1 #include <stdio.h>
 2 #include <string.h>
 3 #define maxn 300
 4 int str1[maxn][maxn];
 5 int main()
 6 {
 7     int T, i, j;
 8     scanf("%d ",&T);
 9 
10     while(T--)
11     {
12         char str[300];
13         gets(str);
14 
15         int len=strlen(str);
16         str1[0][0]=1;
17         str1[1][0]=2;
18 
19         for(i=2;i<len;i++)
20         {
21             int c=0,k;
22             for(k=0;k<maxn;k++)
23             {
24                 int s=str1[i-1][k]+str1[i-2][k]+c;
25                 str1[i][k]=s%10;
26                 c=s/10;
27             }
28             while(c)
29             {
30                 str1[i][k++]=c%10;
31                 c/=10;
32             }
33         }
34 
35         for(i=299;i>=0;i--)
36             if(str1[len-1][i])
37                 break;
38 
39         for(j=i;j>=0;j--)
40             printf("%d",str1[len-1][j]);
41             printf("\n");
42     }
43     return 0;
44 }

A完还A：

 1 #include<stdio.h>
 2 #include<string.h>
 3 
 4 #define N 210
 5  int f[1010][N];
 6 
 7  void init()
 8  {
 9      int i, j;
10      f[0][1] = 1;
11        f[1][1] = 1;
12 
13        for(i = 2; i <= 1000; i++)
14         for(j = 1; j <= 200; j++)
15        {
16            f[i][j] += f[i-1][j] + f[i-2][j];
17            if(f[i][j] >= 10)
18            {
19                f[i][j] -= 10;
20                f[i][j+1]++;
21            }
22        }
23  }
24 
25 int main()
26 {
27 
28    int i, j, t;
29    char s[N];
30 
31    scanf("%d", &t);
32    init();
33 
34 
35    while(t--)
36    {
37        scanf("%s", s);
38 
39        int n = strlen(s);
40 
41 
42         int k = 150;
43         for (; f[n][k] == 0 ; k--);
44         for (; k >= 1 ; k--)
45             printf ("%d",f[n][k]);
46 
47             printf("\n");
48 
49    }
50 
51    return 0;
52 }

posted on 2016-08-04 14:57 惟愿。。。阅读(134) 评论(0) 编辑收藏举报