1054. The Dominant Color (20)
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 通过查找，每次从列表中除去两个不一样的数，最后就可以得出这个数，时间复杂度O（N）。写法上也有技巧，不必非要找到一个不一样的在继续下去，如果下一个一样，那么用一个变量记录这个次数，把次数+1，遇见不一样的－1。例如1，1，2，3，4 初始value ＝1，count ＝1，第二个1，value ＝ 1，count ＝2，下一个2，value＝1，count=1,下一个3，value＝3，count=0,下一个4，value＝4，count＝1...一直到最后，看下value的值，就是所要找的数。
#include"iostream"
#include "algorithm"
#include <string>
#include "cstring"
#include"sstream"
using namespace std;
int main()
{
int m,n;
scanf("%d%d",&m,&n);
int Times=0,value;
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
int num;
scanf("%d",&num);//用cin会超时
if(Times ==0)
{
value = num;
Times++;
}
else
{
if(num == value)
Times++;
else
Times--;
}
}
printf("%d\n",value);
return 0;
}