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; }
时间: 2024-10-14 12:33:00