大体题意：
给你n 个数，要求1~m每个数出现次数的最小值尽可能大，要求你修改n个数，要改的次数尽可能小，输出最小值的最大可能和最少修改次数，并把改的数组输出出来！
思路：
开始没看到次数尽可能少，wa了一次！
想一想就知道，最大可能值肯定是 n/m
那么直接统计1~m 每个数出现了几次！
然后枚举一遍a数组，如果当前值 在1~m范围内！就判断这个数的出现次数 如果小于等于n/m 肯定不用管！
如果大于n/m 那么就枚举一遍1~m 看哪个数 还不足 n/m 就给它加上！
如果不在1~m范围内 同样的方式处理。
时间复杂度 o(nm)
详细见代码：
#include
#define ps push_back
#define fi first
#define se second
using namespace std;
const int maxn = 2000 + 10;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
typedef long long ll;
typedef unsigned long long ULL;
ll a[maxn],b[maxn];
int num[maxn];
int main(){
int n,m;
scanf("%d %d",&n,&m);
for (int i = 1; i 1)printf(" ");
printf("%d",b[i]);
}
puts("");
return 0;
}C. Polycarp at the Radio
time limit per test
memory limit per test
input
output
a1, a2, …, an, where ai is a band, which performs the i-th song. Polycarp likes bands with the numbers from 1 to m, but he doesn’t really like others.
bj the number of songs the group j is going to perform tomorrow. Polycarp wants to change the playlist in such a way that the minimum among the numbers b1, b2, …, bm
bj (1 ≤ j ≤ m), and the minimum number of changes in the playlist Polycarp needs to make to achieve it. One change in the playlist is a replacement of the performer of the i-th song with any other group.
Input
n and m (1 ≤ m ≤ n ≤ 2000).
n integers a1, a2, …, an (1 ≤ ai ≤ 109), where ai is the performer of the i-th song.
Output
bj (1 ≤ j ≤ m), where bj is the number of songs in the changed playlist performed by the j-th band, and the minimum number of changes in the playlist Polycarp needs to make.
In the second line print the changed playlist.
If there are multiple answers, print any of them.
Examples
input
4 2
1 2 3 2
output
2 1
1 2 1 2
input
7 3
1 3 2 2 2 2 1
output
2 1
1 3 3 2 2 2 1
input
4 4
1000000000 100 7 1000000000
output
1 4
1 2 3 4
Note
b1), and the second band also performs two songs (b2). Thus, the minimum of these values equals to 2. It is impossible to achieve a higher minimum value by any changes in the playlist.
b1), the second band performs three songs (b2), and the third band also performs two songs (b3). Thus, the best minimum value is 2.
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