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