POJ 2579 Blurred Vision（我的水题之路——四方格平均值）
Blurred Vision
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 1449 | Accepted: 1053 |
Description
Aliasing is the stair-step effect achieved when attempting to represent a smooth curve using a finite number of discrete pixels. Of course, all computer displays consist of a finite number of pixels, and many strategies have been devised to smooth the jagged edges with varying degrees of success.
Boudreaux and Thibodeaux are writing video game rendering software for the next big first-person shooter, and they don't know much about any of the progress made in the field of anti-aliasing. Therefore, they've decided to use a very simplistic (and visually unappealing) method to smooth the ragged edges. Unfortunately, it blurs the entire image, but at least it gets rid of those jaggies!
Normally, the game displays in m x n pixels, but they perform an extra anti-aliasing step that converts that image into an (m - 1) x (n - 1) image. Nobody will notice a pixel missing from each dimension, and they can calculate the new pixels by averaging squares of 4 pixels from the original image (and rounding down). For example, the images below represent the original image (left) and the anti-aliased image (right) using numbers to represent varying shades of black and white.
Boudreaux and Thibodeaux are writing video game rendering software for the next big first-person shooter, and they don't know much about any of the progress made in the field of anti-aliasing. Therefore, they've decided to use a very simplistic (and visually unappealing) method to smooth the ragged edges. Unfortunately, it blurs the entire image, but at least it gets rid of those jaggies!
Normally, the game displays in m x n pixels, but they perform an extra anti-aliasing step that converts that image into an (m - 1) x (n - 1) image. Nobody will notice a pixel missing from each dimension, and they can calculate the new pixels by averaging squares of 4 pixels from the original image (and rounding down). For example, the images below represent the original image (left) and the anti-aliased image (right) using numbers to represent varying shades of black and white.
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Input
Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.
A single data set has 3 components:
After the last data set, there will be a single line:
ENDOFINPUT
A single data set has 3 components:
- Start line - A single line:
START R C
where R and C are integers (2 <= (R,C) <= 9) indicating the number of rows and columns in the input image described by this data set.
- Original Image - A series of R lines, each of which contains C integers from 0 to 9 inclusive. These integers represent the grayscale value of a pixel in the original image and will not be separated by spaces.
- End line - A single line:
END
After the last data set, there will be a single line:
ENDOFINPUT
Output
The output will be the anti-aliased image, which will be R - 1 rows, each with C - 1 integer pixel values. Each pixel in the output will be generated by averaging (and rounding down) the grayscale pixel values of the corresponding square of four pixels in the Original Image.
Sample Input
START 2 2 00 00 END START 2 9 012345678 012345678 END START 4 4 4440 4400 4000 0000 END START 9 9 900000009 090000090 009000900 000909000 000090000 000909000 009000900 090000090 900000009 END ENDOFINPUT
Sample Output
0 01234567 431 310 100 42000024 24200242 02422420 00244200 00244200 02422420 24200242 42000024
Source
South Central USA 2003
对于一个R*C的矩阵，输出矩阵中每一个2*2的小方格四个值的平均值（向下取整）。如下图：
先将所有元素读入在一个二维数组中，[0,0] ~[n-1, n-1]。
然后从[1,1]~[n-1, n-1]，取[i,j]元素和[i-1,j]、[i,j-1]、[i-1,j-1]元素的平均值，输出。
注意点：
1）输入值前有“STRAT”，之后有“END”字符串，需要忽略这些输入。
2）这些数字中间没有空格间隔，需要用字符读入，注意读入之后的'\n'影响下一步读入。
3）取中间值，所以注意边界，不要越界。
代码（1AC）：
#include <cstdio>
#include <cstdlib>
#include <cstring>
int arr[10][10];
char sign[3][20] = {"START", "END", "ENDOFINPUT"};
int main(void){
int i, j;
int r, c;
int ave;
char tmp[20];
while (scanf("%s", tmp), strcmp(tmp, sign[2])){
memset(arr, 0, sizeof(arr));
scanf("%d%d", &r, &c);
getchar();
for (i = 0; i < r; i++){
for (j = 0; j < c; j++){
arr[i][j] = getchar() - '0';
}
getchar();
}
for (i = 1; i < r; i++){
for (j = 1; j < c; j++){
ave = arr[i][j] + arr[i - 1][j] + arr[i][j - 1] + arr[i - 1][j - 1];
printf("%d", ave / 4);
}
printf("\n");
}
scanf("%s", tmp);
getchar();
}
return 0;
}