1054. The Dominant Color (20)-PAT

1054. The Dominant Color (20)

时间限制

100 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

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.

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

推荐指数：※※

来源：http://pat.zju.edu.cn/contests/pat-a-practise/1054

这道题目想到了方法就很简单，正好，我看过编程之美2.3寻找发帖”水王“，这一题只是壳子不同，输入的时候不相同的互相抵消，多余一半的肯定不会被抵消完的。通俗的来说：有一个椅子，有很多teams都想坐上去，每个team各个队员的战斗力是一样的（不同team队员遇到1:1消失），同样的成员都会在椅子上。现在所有人混在一起，排成一大队一个一个的去抢椅子，如果有一个team的队员大于总队员一半，1:1战斗完后剩下的一定是他的队员。

不然用map也可以。

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
	int m,n;
	scanf("%d%d",&m,&n);
	int i,j,candidate,ntimes=0;
	for(i=0;i<m;i++){
		for(j=0;j<n;j++)
		{
			int color;
			scanf("%d",&color);
			if(ntimes==0){
				candidate=color;
				ntimes++;
			}
			else
			{
				if(candidate==color)
					ntimes++;
				else
					ntimes--;
			}
		}
	}
	printf("%d\n",candidate);
	return 0;
}