CODE FESTIVAL 2017 qual A C

C - Palindromic Matrix
Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement
We have an H-by-W matrix. Let aij be the element at the i-th row from the top and j-th column from the left. In this matrix, each aij is a lowercase English letter.

Snuke is creating another H-by-W matrix, A’, by freely rearranging the elements in A. Here, he wants to satisfy the following condition:

Every row and column in A’ can be read as a palindrome.
Determine whether he can create a matrix satisfying the condition.

Note
A palindrome is a string that reads the same forward and backward. For example, a, aa, abba and abcba are all palindromes, while ab, abab and abcda are not.

Constraints
1≤H,W≤100
aij is a lowercase English letter.
Input
Input is given from Standard Input in the following format:

H W
a11a12…a1W
:
aH1aH2…aHW
Output
If Snuke can create a matrix satisfying the condition, print Yes; otherwise, print No.

Sample Input 1
Copy
3 4
aabb
aabb
aacc
Sample Output 1
Copy
Yes
For example, the following matrix satisfies the condition.

abba
acca
abba
Sample Input 2
Copy
2 2
aa
bb
Sample Output 2
Copy
No
It is not possible to create a matrix satisfying the condition, no matter how we rearrange the elements in A.

Sample Input 3
Copy
5 1
t
w
e
e
t
Sample Output 3
Copy
Yes
For example, the following matrix satisfies the condition.

t
e
w
e
t
Sample Input 4
Copy
2 5
abxba
abyba
Sample Output 4
Copy
No
Sample Input 5
Copy
1 1
z
Sample Output 5
Copy
Yes
思维，数学

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>#include <string>using namespace std;char str[111][111];int a[30];int main(){    int h,w;    scanf("%d %d",&h,&w);    for(int i=0;i<h;i++){        scanf("%s",str[i]);        for(int j=0;j<w;j++){            a[str[i][j]-'a'+1]++;        }    }    int g1=(h%2)*(w%2);    int g2=(h/2)*(w%2)+(w/2)*(h%2);    int g3=(h/2)*(w/2);    for(int i=1;i<=26;i++){        g3-=(a[i]/4);        a[i]%=4;        g2-=(a[i]/2);        a[i]%=2;        g1-=a[i];    }    g2+=(g3*2);    if(g3>0||g2>0||g1>0){        puts("No");        return 0;    }    puts("Yes");    return 0;}