The Dominant Color (20)

题目描述

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, 800x600), you are supposed to point out the strictly dominant color.

输入描述:

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.

输出描述:

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

输入例子:

5 3
0 0 255 16777215 24
24 24 0 0 24
24 0 24 24 24

输出例子:

24

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <fstream>
#include <cmath>
#include <string>
/*
这题就是找出出现最多次的一个数字而已，但是有一句比较重要的信息，就是这个数字比总数字的数量一半还多，
比如：100个数字，这个数字至少有51个。 所以应用相同累加，不同删除的方法，一次遍历即可把这个数字找出
来。 复杂度只有O(n)
*/

//#define LOCAL
using namespace std;

int main()
{
#ifdef LOCAL
#define cin fin
ifstream fin("test.txt");
#endif // LOCAL
int m, n;				//行、列变量
int maxNum = 0;			//保存当前相同累加，不同删除所得到的最大出现次数
long long last = -1;	//当前保存的没有删除的数字
bool saveLast = true;	//如果删除掉所有数字在存入一个新的
cin >> n >> m;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
{
//因为是2^24，所以需要用long long型
long long temp;
cin >> temp;
if (saveLast)
{
last = temp;
maxNum++;
saveLast = false;
continue;
}
//相同累加，不同删除
if (temp == last)
{
maxNum++;
}
else if(--maxNum == 0)
{
//没有比较的数字了，重新载入新的数字
saveLast = true;
}
}
cout << last;
//getchar();
}