2021.5.22比赛补题

本次比赛发挥较好，完成了ABC三个题，这对自己是个鼓励，望下继续努力，其实cf的题并不是难，代码操作量一般不大，更多是对思维的训练。比的就是对问题本质发现速度。
补一下c题：
One day little Vasya found mom’s pocket book. The book had n names of her friends and unusually enough, each name was exactly m letters 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).

Examples
Input
2 3
AAB
BAA
Output
4
Input
4 5
ABABA
BCGDG
AAAAA
YABSA
Output
216
Note
In the first sample Vasya can get the following names in the position number 1: “AAB”, “AAA”, “BAA” and “BAB”.
题解：c题一开始看起来很难，但关键就是找到问题本质，字符串前缀可以交换次数不限，这样就直接求每一列有多少个不同字母，然后每一列相乘即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll nl=1e2+5;
string a[nl];
ll b[nl]={0};
ll c[nl]={0};
ll ml=1e9+7;
int main(){
	ll n,m;
	ll i,j;
	cin>>n>>m;
	for(i=0;i<n;i++){
		cin>>a[i];
		
	}
	for(i=0;i<m;i++){
		for(j=0;j<n;j++){
			b[a[j][i]-'A']=1;//字符串变成数字常用方式
		}
		for(j=0;j<26;j++){
			if(b[j]>0){
				c[i]++; 
			} 
		}
		for(j=0;j<26;j++){
			b[j]=0;
		}
	}
	ll sum=1;
	for(i=0;i<m;i++){
		sum=sum*c[i]%ml;//注意每一次都要取模
	}
	cout<<sum;
}