PAT甲级刷题记录——1054 The Dominant Color (20分)

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

思路

这题很简单哈，其实如果范围不大的话，用数组hash来做也行，但是这个给的数范围太大了（2²⁴），那么只好用map来做了。

题意很简单，所谓主色（“dominant color”）就是出现次数最多的那个颜色，以样例来说，颜色0出现了5次，颜色24出现了8次，颜色255出现了1次，颜色16777215出现了1次，因此，颜色24出现的次数最多，所以它是主色，输出24。

所以，只要用map存储完所有数据之后，用迭代器从头到尾找一下出现次数最多的颜色就行啦~

代码

#include
#include
#include
#include
#include
#include
using namespace std;
map<int, int> color;
int main(){
    int M, N;
    scanf("%d%d",&M, &N);
    for(int i=0;i<N;i++){
        for(int j=0;j<M;j++){
            int tmp;
            scanf("%d", &tmp);
            color[tmp]++;
        }
    }
    int DominantColor = 0;
    int cnt = 0;
    for(map<int, int>::iterator it=color.begin();it!=color.end();it++){
        if(it->second>cnt){
            DominantColor = it->first;
            cnt = it->second;
        }
    }
    printf("%d", DominantColor);
    return 0;
}