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
题意:
给出 M 列 N 行的数字矩阵,求其中超过半数的出现次数最多的数字
输入:M (≤800) and N (≤600)
思路:
首先想到的是采用数组存储,进行计数即可。但是忽略了一点,给出的数据很可能很大,导致内存超限
因此采用 map ,建立 map<int, int> ,作为数字与其出现次数的映射关系
知识点:
map
- find(key) 返回键为 key 的映射的迭代器
- map 的访问有两种方式:下标和迭代器。迭代器访问的定义方式为: map<typename1, typename2>::iterator it; 其中,采用 it->first 访问键,采用 it->second 访问值
#include <cstdio>
#include <map>
using namespace std;
int main(){
int n, m, num;
scanf("%d%d", &n, &m);
map<int, int> count;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
scanf("%d", &num);
if(count.find(num) != count.end())
count[num]++;
else
count[num] = 1;
}
}
int k, max = 0;
for(map<int, int>::iterator it = count.begin(); it != count.end(); it++){
if(it->second > max){
k = it->first;
max = it->second;
}
}
printf("%d\n", k);
return 0;
}