题意:给 n 个数,输出众数,但是如果所有的频率都相同但数不同输出 Bad Mushroom。
析:直接记录个数直接暴力就,就是要注意只有一种频率的时候。
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 <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 10000 + 10; const int mod = 100000000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; 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 bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[maxn]; int main(){ int T; cin >> T; for(int kase = 1; kase <= T; ++kase){ scanf("%d", &n); memset(a, 0, sizeof a); for(int i = 0; i < n; ++i){ int x; scanf("%d", &x); x = 10000 - (100-x) * (100-x); ++a[x]; } int mmax = 0; vector<int> v; int cnt = 0; for(int i = 0; i <= 10000; ++i){ if(a[i] == 0) continue; ++cnt; if(mmax < a[i]){ v.clear(); mmax = a[i]; v.push_back(i); } else if(mmax == a[i]) v.push_back(i); } printf("Case #%d:\n", kase); if(cnt == 1) printf("%d", v[0]); else if(cnt == v.size()) printf("Bad Mushroom"); else for(int i = 0; i < v.size(); ++i){ if(i) putchar(' '); printf("%d", v[i]); } printf("\n"); } return 0; }