每日一题之 hiho207 The Lastest Time

描述
What is latest time you can make with 4 digits A, B, C and D?

For example if the 4 digits are 1, 0, 0, 0, you can make 4 times with them: 00:01, 00:10, 01:00, 10:00. The lastest time will be 10:00. Note a valid time is between 00:00 and 23:59.

输入
One line with 4 digits A, B, C and D, separated by a space. (0 <= A, B, C, D <= 9)

输出
Output the lastest time in the format “hh:mm”. If there is no valid time output NOT POSSIBLE.

样例输入
0 9 0 0
样例输出
09:00

思路：

由于只有四个数字，那么我们可以直接枚举这四个数的排列。然后将前两两位看作是hours和后两位数看做是minutes，从所有排列中选择一个符合要求并且最晚的时间输出就行

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <string>

using namespace std;

bool nextPermutation(string &str,int num) {
    int i;
    for ( i = num - 2; (i >= 0) && (str[i] >= str[i+1]); --i);
    if ( i < 0) return false;
    int k;
    for ( k = num - 1; (k > i) && (str[k] <= str[i]); --k);
    swap(str[i],str[k]);
    reverse(str.begin()+i+1,str.end());
    return true;
}

int main()
{
    string str = "",x;
    for (int i = 0; i < 4; ++i) {
        cin >> x;
        str += x;
    }
    int hour = -1,minutes = -1;
    sort(str.begin(),str.end());

   do {
        int tmp_h = (str[0]-'0')*10 + str[1]-'0';
        int tmp_m = (str[2]-'0')*10 + str[3]-'0';
        if (tmp_h >= 0 && tmp_h <= 23 && tmp_m >= 0 && tmp_m <= 59) {
                if (tmp_h > hour || (tmp_h == hour && tmp_m > minutes)) {
                    hour = tmp_h;
                    minutes = tmp_m;
                }
        }

    }while(nextPermutation(str,4));

    if (hour == -1 || minutes == -1) cout << "NOT POSSIBLE" <<endl;
    else {
        if (hour < 10) cout << "0" << hour;
        else
            cout << hour;
        cout << ":";
        if (minutes < 10) cout << "0" << minutes;
        else
            cout << minutes;
        cout << endl;
    }

    return 0;
}