HDU 1029 Basic DP

Title Link: Ignatius and the Princess IV

The main idea is to find the only number in the N number that appears at least (n+1)/2 times. 1 <= N <= 999999.

Hash

1 /*2 Test instructions understood. is to find the only number in the N number that appears at least (n+1)/2 times. 1 <= N <= 999999.3 If I use a map tag. Maybe it's just 10^6. 4 However, it is still stuck with CIN and cout. 5  */6 7#include <stdio.h>8#include <string.h>9#include <iostream>Ten#include <map> One using namespacestd; A  -map<int,int>MP; -  the intMain () { -     intN; -     intnum; -     intans; +  -      while(~SCANF ("%d", &N)) { + mp.clear (); A          for(intI=0; i<n; ++i) { atscanf"%d", &num); -mp[num]++; -             if(Mp[num] >= (n+1)/2) {//If you find it, you can't break it. Because you may not have finished typing, -Ans =num; -             } -         } inprintf"%d\n", ans); -     } to     return 0; +}

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Dp:

1 /*2 get to a very ingenious method. Follow test instructions. The number I asked for was the most frequently seen in the numbers,3 Compare the number of occurrences of the current number with the most number in all the preceding numbers at a time. 4 Judging from the process? is the dynamic planning of this problem? 5  */6 7#include <stdio.h>8#include <string.h>9#include <string.h>Ten#include <iostream> One using namespacestd; A  - intMain () { -     intN; the     intnum, CNT, REM; -      while(~SCANF ("%d", &N)) { -CNT =0; -          for(intI=0; i<n; ++i) { +scanf"%d", &num); -             if(CNT = =0) {//Count +REM =num; ACNT =1; at             } -             Else if(num = =REM) { -cnt++; -             } -             Else { -cnt--; in             } -         } toprintf"%d\n", REM); +     } -     return 0; the}

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HDU 1029 Basic DP