CodeForces 152C Pocket Book

最新推荐文章于 2021-05-22 16:52:40 发布

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最新推荐文章于 2021-05-22 16:52:40 发布

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原文链接：http://www.cnblogs.com/cyd308/p/4771491.html

本博客介绍了一个Codeforces挑战题目的解决方案，该题目涉及字符串操作和数学计算，玩家Vasya通过特定规则交换名字的前缀，最终计算在不同操作后名字在口袋书中可能的位置变化。

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Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 152C

Description

One day little Vasya found mom's pocket book. The book had n names of her friends and unusually enough, each name was exactly mletters long. Let's number the names from 1 to n in the order in which they are written.

As mom wasn't home, Vasya decided to play with names: he chose three integers i, j, k (1 ≤ i < j ≤ n, 1 ≤ k ≤ m), then he took names number i and j and swapped their prefixes of length k. For example, if we take names "CBDAD" and "AABRD" and swap their prefixes with the length of 3, the result will be names "AABAD" and "CBDRD".

You wonder how many different names Vasya can write instead of name number 1, if Vasya is allowed to perform any number of the described actions. As Vasya performs each action, he chooses numbers i, j, k independently from the previous moves and his choice is based entirely on his will. The sought number can be very large, so you should only find it modulo 1000000007(10⁹ + 7).

Input

The first input line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of names and the length of each name, correspondingly. Then n lines contain names, each name consists of exactly m uppercase Latin letters.

Output

Print the single number — the number of different names that could end up in position number 1 in the pocket book after the applying the procedures described above. Print the number modulo 1000000007(10⁹ + 7).

Sample Input

Input

2 3
AAB
BAA

Output

Input

4 5
ABABA
BCGDG
AAAAA
YABSA

Output

Hint

In the first sample Vasya can get the following names in the position number 1: "AAB", "AAA", "BAA" and "BAB".

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int n,m;
 8     int i,j,k;
 9     char a[105][105];
10     int b[30];
11     long long c[1005];
12     while(scanf("%d %d",&n,&m)!=EOF)
13     {
14         memset(c,0,sizeof(c));
15         for(i=1;i<=n;i++)
16             scanf("%s",a[i]);
17         for(i=0;i<m;i++)
18         {
19             memset(b,0,sizeof(b));
20             for(j=1;j<=n;j++)
21             {
22                 b[a[j][i]-'A']++;
23             }
24             for(j=0;j<=28;j++)
25                 if(b[j])
26                     c[i]++;
27             //printf("%I64d\n",c[i]);
28         }
29         long long ans=c[0];
30         for(i=1;i<m;i++)
31             ans=ans*(c[i]%1000000007)%1000000007;
32         printf("%I64d\n",ans);
33     }
34     return 0;
35 }