PAT甲级—1054 The Dominant Color
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 [0,2
24
). 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
#include<iostream>
using namespace std;
const int MAXN=16777216;
int main(){
int m,n;
while(scanf("%d%d",&m,&n)!=EOF){
int a[n][m];
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
scanf("%d",&a[i][j]);
}
}
long long b[MAXN]={
0};
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
b[a[i][j]]++;
}
}
int num=b[0];
int k=0;
for(int i=0;i<MAXN;i++){
if(num<b[i]){
num=b[i];
k=i;
}
}
printf("%d",k);
}
return 0;
}
错误
参考答案
#include <iostream>
#include <map>
using namespace std;
int main() {
int m, n;
scanf("%d %d", &m, &n);
map<int, int> arr;
int half = m * n / 2;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
int temp;
scanf("%d", &temp);
arr[temp]++;
if(arr[temp] > half) {
printf("%d", temp);
return 0;
}
}
}
return 0;
}