CodeForces 723C Polycarp at the Radio (题意题+暴力)

题意：给定 n 个数，让把某一些变成 1-m之间的数，要改变最少，使得1-m中每个数中出现次数最少的尽量大。

析：这个题差不多读了一个小时吧，实在看不懂什么意思，其实并不难，直接暴力就好，n m不大。很明显最后1-m中次数最长的应该是n/m，

所以我们把大于n/m的都变成小于等于的，把这 n 个数中大于 m 的也变成，但是并不需要都变，只要满足每个数都是大于等于n/m就好了。

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 <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e3 + 5;
const LL mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn], b[maxn];

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        memset(b, 0, sizeof b);
        for(int i = 1; i <= n; ++i){
            scanf("%d", a+i);
            if(a[i] <= m)  ++b[a[i]];
        }

        int ans1 = n / m;
        int ans2 = 0;

        int cnt = 1;
        for(int i = 1; i <= n; ++i) if(a[i] <= m && b[a[i]] > ans1){
            --b[a[i]];
            for(int j = cnt; j <= m; ++j){
                if(b[j] < ans1) {  ++b[j]; cnt = j; a[i] = j; ++ans2; break; }
            }
        }else if(a[i] > m){
            for(int j = cnt; j <= m; ++j){
                if(b[j] < ans1) {  ++b[j]; cnt = j; a[i] = j; ++ans2; break; }
            }
        }

        printf("%d %d\n", ans1, ans2);
        for(int i = 1; i <= n; ++i)
            if(i == 1)  printf("%d", a[i]);
            else printf(" %d", a[i]);
        printf("\n");
    }
    return 0;
}