PAT-A-1054 The Dominant Color

Behind the scenes in the computer’s memory, color is always talked about as a series of 24 bits of information for each pixel. In an image, the color with the largest proportional area is called the dominant color. A strictly dominant color takes more than half of the total area. Now given an image of resolution M by N (for example, 800×600), you are supposed to point out the strictly dominant color.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: M (≤800) and N (≤600) which are the resolutions of the image. Then N lines follow, each contains M digital colors in the range [0,2

​24

​​ ). It is guaranteed that the strictly dominant color exists for each input image. All the numbers in a line are separated by a space.

Output Specification:

For each test case, simply print the dominant color in a line.

Sample Input:

5 3

0 0 255 16777215 24

24 24 0 0 24

24 0 24 24 24

Sample Output:

24

方法1：用map來求解獎勵map<int,int> cnt 作為數字與出現次數的映射關系

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <string.h>
#include <math.h>
using namespace std;

int main(){
    int n,m,col;
    map<int,int> cnt;
    scanf("%d%d",&m,&n);
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            scanf("%d",&col);
            if(cnt.find(col)!=cnt.end()){
                cnt[col]++;//若已存在就次數加一
            }else{
                cnt[col]=1;//若不存在次數置為1
            }
        }
    }
    int k=0,MAX=0;//最大數字及該數字出現次數
    for(map<int,int>::iterator it=cnt.begin();it!=cnt.end();it++){
        if(it->second>MAX){
            k=it->first;
            MAX=it->second;
        }
    }
    printf("%d\n",k);
    // system("pause");
    return 0;
}
           

方法2：設定cnt和ans,cnt負責計數，ans負責記錄答案,遇到相等的cnt加1，遇到不等的cnt減1，當cnt變為0時，将ans換為新的數，cnt重置為1

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <string.h>
#include <math.h>
using namespace std;

int main()
{
    int n, m, col;
    scanf("%d%d", &m, &n);
    int ans, cnt; //ans為衆數，cnt為這個數出現的次數
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            scanf("%d", &col);
            if (i == 0 && j == 0)
            {
                cnt = 1;
                ans = col;
            }
            else
            {
                if (col == ans)//如果依舊是這個數
                {
                    cnt++;
                }
                else//如果不是這個數
                {
                    cnt--;
                    if (cnt == 0)
                    { //如果cnt為0令新數字為ans
                        ans = col;
                        cnt++; //cnt重新置為1
                    }
                }
            }
        }
    }
    printf("%d\n",ans);
    // system("pause");
    return 0;
}