1. 题目
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 [0,224). 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
2. 题意
给出m×n大小的屏幕，每个点位会有一个颜色值，找出整个屏幕中颜色值占最多那个颜色。
3. 思路——map的使用
创建一个map，键值对为<颜色, 数量>,对屏幕中的所有颜色进行计数，计数完成后，遍历找出数量最多的那个颜色。
4. 代码
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
int m, n;
cin >> m >> n;
map<int, int> res;
for (int i = 0; i < m; ++i)
{
for (int j = 0; j < n; ++j)
{
int x;
cin >> x;
res[x] += 1;
}
}
int dominantColor = 0;
int maxCnt = 0;
// map的遍历
// 如果有新颜色在屏幕中出现次数大于当前记录的颜色，则更新
for (map<int, int>::iterator iter = res.begin(); iter != res.end(); iter++)
if (iter->second > maxCnt) maxCnt = iter->second, dominantColor = iter->first;
cout << dominantColor << endl;
return 0;
}