The Dominant Color (20)

时间限制 1000 ms 内存限制 65536 KB 代码长度限制 100 KB 判断程序 Standard (来自 小小)

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();
}