题意：给定 n 和 k，然后是 n 个数，表示1-k的一个值，问你修改最少的数，使得所有的1-k的数目都等于n/k。
析：水题，只要用每个数减去n/k，然后取模，加起来除以2，就ok了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
int n, m;
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[105];
int main(){
while(scanf("%d %d", &n, &m) == 2){
memset(a, 0, sizeof(a));
for(int i = 0; i < n; ++i){
int x;
scanf("%d", &x);
++a[x];
}
int x = n / m;
int ans = 0;
for(int i = 1; i <= m; ++i)
ans += abs(x-a[i]);
printf("%d
",ans/2);
}
return 0;
}