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:
", 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("
");
}
return 0;
}