时间限制：100ms
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
==============================
这题目就是个投票选班长的题目，上个学期数据库老师课上讲过（可惜了，数据库后来没考好），瞬间犹如醍醐灌顶~
醍醐灌顶，出于《敦煌变文集·维摩诘经讲经文》：“令问维摩，闻名之如露入心，共语似醍醐灌顶。”佛教指灌输智慧，使人彻底觉悟。比喻听了高明的意见使人受到很大启发。
最笨的办法是每个人画正字，浪费空间，最后还要统计票数
最终的目的是要选出票数最多的一个人，那么可以这样做：
初始化:
班长候选人：叉叉叉
当前差额票数：1
然后拿一张选票，票面上的名字是Name
if Name == 当前的候选人:
差额票数++
else:
差额票数--
if 差额票数==0:
候选人 = Name
差额票数 = 1
这样的话，省时间也省空间:)
原理其实想通了很简单，因为我们只选一个人啊，只算净胜票数就行
净胜票数为0了，那就要被赶下台了！
从测试点的结果来看，的确有个测试点数据量挺大的…估计之前说的笨办法就要超时了..
=================================
#include
int main()
{
int M, N;
scanf("%d %d", &M, &N);
int resolution = M * N;
int currBest = -1;
int currCount = 1;
for(int i=0; i
测试点 结果 用时(ms) 内存(kB) 得分/满分 0 答案正确 0 790 12/12 1 答案正确 0 790 2/2 2 答案正确 60 780 2/2 3 答案正确 0 790 2/2 4 答案正确 0 710 2/2