HDU 1800 Flying to the Mars detailed hash

Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1800

This topic is Dhahashi's simple question, the main difficulty lies in how to abstract the question painting.

For each set of data, I require a minimum of a few brooms.

We sort 2 4 5 6 4 for this set of inputs, become 2 4 4 5 6, each time take the longest increment sequence, take the number of times is the answer we need, please think carefully, if the input is 2 4 5 6, then we only need a broom to install everyone, but we enter 2 4 4 5 6, we need two, we each have taken an increment sequence, why do we start with one, and now we need two? Because I also have a person's rank is also 4, and I already have a 4 and the sequence must increment, so the same level of 4 I can only take one person. The other had to keep the next time to fetch. So the question is abstract: Ask for the number of times the most frequent occurrence of a set of numbers!

Is it easy to think like this?

Each time you enter a number I, the array mp[i]++ once, then the scope of I all over again, the MP array of the largest element output on the line. (If you do not know the map inside the STL can own Baidu a bit)

Put in a complete code:

#include <iostream> #include <string> #include <stdio.h> #include <map>using namespace Std;int Main () {    int n,max;    map<int,int>mp;    int num;    while (scanf ("%d", &n)!=eof) {        mp.clear ();        max=0;        while (n--) {            scanf ("%d", &num);            mp[num]++;            if (Mp[num]>max) {                max=mp[num];            }        }        printf ("%d\n", max);    }    return 0;}

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HDU 1800 Flying to the Mars detailed hash