1054. The Dominant Color (20)-PAT甲级真题

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

题目大意：选取主色调，就是M列N行的矩阵里面出现次数多余一半的那个数字～

分析：STL中map的应用～使用arr[i] = j表示i元素在矩阵中出现了j次，在输入的同时比较arr当前的值是否已经超过半数，如果超过，就直接输出该数字并退出程序～

#include <iostream>
#include <map>
using namespace std;
int main() {
    int m, n;
    scanf("%d %d", &m, &n);
    map<int, int> arr;
    int half = m * n / 2;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            int temp;
            scanf("%d", &temp);
            arr[temp]++;
            if(arr[temp] > half) {
                printf("%d", temp);
                return 0;
            }
        }
    }
    return 0;
}

#include <iostream>

#include <map>

using namespace std;

int main() {

int m, n;

scanf("%d %d", &m, &n);

map<int, int> arr;

int half = m * n / 2;

for(int i = 0; i < n; i++) {

for(int j = 0; j < m; j++) {

int temp;

scanf("%d", &temp);

arr[temp]++;

if(arr[temp] > half) {

printf("%d", temp);

return 0;

}

return 0;

}

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