Gym 100531A Alarm Clock (水题)

题意：给定一个被高亮的数，问你是不是有个时间恰好高亮是这个数。

析：直接暴力，直接暴力，枚举每一位时间，当然也可以枚举时间，能找到就是有，找不到就算了。

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>
#define debug puts("+++++")
//#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 = 200 + 5;
const LL mod = 1e3 + 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); }
inline int lcm(int a, int b){  return a * b / gcd(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;
}
map<int, int> mp;

int main(){
    freopen("alarm.in", "r", stdin);
    freopen("alarm.out", "w", stdout);
    mp[0] = 6, mp[1] = 2, mp[2] = 5, mp[3] = 5, mp[4] = 4;
    mp[5] = 5, mp[6] = 6, mp[7] = 3, mp[8] = 7, mp[9] = 6;
    while(scanf("%d", &n) == 1){
        bool ok = false;
        int h, t;
        for(int i = 0; i < 3 && !ok; ++i)
            for(int j = 0; j < 10 && !ok; ++j)
                for(int k = 0; k < 6 && !ok; ++k)
                    for(int l = 0; l < 10 && !ok; ++l)
                        if(i*10 + j < 24 && k*10 + k < 60 && mp[i] + mp[j] + mp[k] + mp[l] == n){
                            ok = true;
                            h = i * 10 + j;
                            m = k * 10 + l;
                        }
        if(ok)  printf("%02d:%02d\n", h, m);
        else puts("Impossible");
    }
    return 0;
}