There are nn students in the class. The test had mm questions, each of them had 55 possible answers (A, B, C, D or E). There is exactly one correct answer for each question. The correct answer for question ii worth aiai points. Incorrect answers are graded with zero points.
The students remember what answers they gave on the exam, but they don't know what are the correct answers. They are very optimistic, so they want to know what is the maximum possible total score of all students in the class.
Input
The first line contains integers nn and mm (1≤n,m≤1000) — the number of students in the class and the number of questions in the test.
Each of the next nn lines contains string sisi (|si|=m), describing an answer of the ii-th student. The jj-th character represents the student answer (A, B, C, D or E) on the jj-th question.
The last line contains mm integers a1,a2,…,am (1≤ai≤1000) — the number of points for the correct answer for every question.
Output
Print a single integer — the maximum possible total score of the class.
Examples
input
2 4
ABCD
ABCE
1 2 3 4
output
16
input
3 3
ABC
BCD
CDE
5 4 12
output
21
Note
In the first example, one of the most optimal test answers is "ABCD", this way the total number of points will be 1616.
In the second example, one of the most optimal test answers is "CCC", this way each question will be answered by exactly one student and the total number of points is 5+4+12=215+4+12=21.
题目，分析就是找每个答案出现最多的那个，然后把把次数最多的相乘，最后累加起来。
听懂了记得给个赞鼓励一下，码字不易，用爱发电。
上ac代码。
有事你就q我；QQ2917366383
学习算法
#include<iostream> #include<algorithm> #include<cstring> using namespace std; const int maxn=1e6+5; #define ll long long int maxm; int main() { int a[2005],b[6]; string s[2005]; int n,m,i=1,j; ll ans=0; cin>>n>>m; while(i<=n) { cin>>s[i]; i++; } i=1; while(i<=m) { cin>>a[i]; i++; } i=1; while(i<=m) { memset(b,0,sizeof(b)); int maxm=-1; j=1; while(j<=n) { b[s[j][i-1]-'A'+1]++; j++; } j=1; while(j<6) { maxm=max(maxm,b[j]); j++; } ans+=maxm*a[i]; i++; } cout<<ans<<endl; }
