Pat1054代码

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.

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:

AC代码：（一）

#include<cstdio>
#include<map>using namespace std;int main(int argc,char *argv[])
{int i,j;int m,n;map<int,int> v;scanf("%d%d",&m,&n);for(i=0;i<n;i++)for(j=0;j<m;j++){int temp;scanf("%d",&temp);v[temp]++;}map<int,int>::iterator it;for(it=v.begin();it!=v.end();it++)if(it->second>m*n/2)break;printf("%d\n",it->first);return 0;
}

其实我觉得可以把本题转换成求众数的问题，不知道为什么最后一个case错误。还是贴出来以后再看吧！为什么？？？

#include<cstdio>
#include<vector>
#include<algorithm>using namespace std;int main(int argc,char *argv[])
{int i,j,len;int ans,temp;vector<int> v;int n,m;scanf("%d%d",&m,&n);for(i=0;i<n;i++){for(j=0;j<m;j++){scanf("%d",&temp);v.push_back(temp);}}sort(v.begin(),v.end());len=1;for(i=1;i<m*n;i++)if(v[i]==v[i-len]){len++;ans=v[i];if(len>n*m/2)break;}printf("%d\n",ans);return 0;
}

AC代码：（二）

#include<cstdio>using namespace std;int main(int argc,char *argv[])
{int m,n;int i,j;int ans,count=0;scanf("%d%d",&m,&n);for(i=0;i<n;i++){for(j=0;j<m;j++){int temp;scanf("%d",&temp);if(count==0){ans=temp;count=1;}else{if(ans==temp)count++;elsecount--;}}}printf("%d\n",ans);return 0;
}

因为该数出现的次数超过一半，所以遇到不相同的数则计数减一，最后剩下的数肯定是出现次数最多的数。。。

AC代码：（三）因为该数出现的次数超过一半，所以排序后中位数一定是它。。。

#include<cstdio>
#include<vector>
#include<algorithm>using namespace std;int main(int argc,char *argv[])
{int i,j,len;int temp;vector<int> v;int n,m;scanf("%d%d",&m,&n);for(i=0;i<n;i++){for(j=0;j<m;j++){scanf("%d",&temp);v.push_back(temp);}}sort(v.begin(),v.end());printf("%d\n",v[(m*n)/2]);return 0;
}

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Pat1054代码

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