Subjects PAT B brush road image filter 1066 (15 minutes)

Image filter 1066 (15 minutes)

The image is the filtered image pixels are unimportant dyed background color, so that the important part is highlighted. Now given a monochrome image, the gray value is located in claim you all pixels within a designated color range are replaced with one specified color.

Input format:
input a given resolution image in the first row, i.e., two positive integers M and N (0 <M, N≤500) , the other is to be filtered gray value interval endpoints A and B (0 ≤A <B≤255), and replacing the tone value specified. Subsequently M rows, each row of the N gradation values given pixel, separated by a space therebetween. All the gradation values are within [0, 255] interval.

Output format:
image output filtering required. I.e. M output lines, each of the N pixel grayscale values, each representing three gradation values (e.g. 000 to be displayed in black), separated by a space therebetween. Line from beginning to end may not have the extra space.

Sample input:
. 3. 5 0 100 150
. 3 189 254 119 101
150 151 99 233 100
88 123 149 0 255

Output Sample:
003.189254 trillion
000 233 151 099 000
088 000 000 000 255

#include <iostream>
using namespace std;

int main()
{
	int M,N,A,B,gray;
	cin>>M>>N>>A>>B>>gray;
	int i,j,point[M][N];
	for(i=0;i<M;i++){
		for(j=0;j<N;j++){
			cin>>point[i][j];
			if(point[i][j]>=A&&point[i][j]<=B){
				point[i][j]=gray;
			}
		}
	}
	for(i=0;i<M;i++){
		for(j=0;j<N;j++){
			printf("%03d",point[i][j]);
			if(j<N-1){
				printf(" ");
			}
		}
		printf("\n");
	}
}