題目：
Color Me Less
Time Limit: 1 Second Memory Limit: 32768 KB
Problem
A color reduction is a mapping from a set of discrete colors to a smaller one. The solution to this problem requires that you perform just such a mapping in a standard twenty-four bit RGB color space. The input consists of a target set of sixteen RGB color values, and a collection of arbitrary RGB colors to be mapped to their closest color in the target set. For our purposes, an RGB color is defined as an ordered triple (R,G,B) where each value of the triple is an integer from 0 to 255. The distance between two colors is defined as the Euclidean distance between two three-dimensional points. That is, given two colors (R1,G1,B1) and (R2,G2,B2), their distance D is given by the equation
The input file is a list of RGB colors, one color per line, specified as three integers from 0 to 255 delimited by a single space. The first sixteen colors form the target set of colors to which the remaining colors will be mapped. The input is terminated by a line containing three -1 values.
Output
For each color to be mapped, output the color and its nearest color from the target set.
Example
Input
0 0 0
255 255 255
0 0 1
1 1 1
128 0 0
0 128 0
128 128 0
0 0 128
126 168 9
35 86 34
133 41 193
128 0 128
0 128 128
128 128 128
255 0 0
0 1 0
0 0 0
255 255 255
253 254 255
77 79 134
81 218 0
-1 -1 -1
Output
(0,0,0) maps to (0,0,0)
(255,255,255) maps to (255,255,255)
(253,254,255) maps to (255,255,255)
(77,79,134) maps to (128,128,128)
(81,218,0) maps to (126,168,9)
Source: Greater New York 2001
題目大意：輸入前16行爲目標顏色組，是一個有序三元組，之後輸入多行顏色組，如果是三個-1，表示結束，目標顏色組的三個顏色設爲r1,g1,b1,其餘的顏色設爲r2,g2,b2,兩者之間的距離公式爲d=sqrt(pow(r2-r1,2)+pow(g2-g1,2)+pow(b2-b1,2))，輸出與後輸入顏色組距離最接近的目標顏色組中的顏色，。
算法分析：其實本題就是求後輸入的顏色與目標顏色組中d最小的顏色，照着題目中的公式計算就可以了。我的方法就是先求出16個目標顏色組中的顏色與輸入顏色的d，在求出最小的d，最後找出最小的d對應的目標顏色。
代碼：
語言：c
#include<stdio.h>
#include<math.h>
int main()
{
int a[16][3],colour[3];
float min,distant[16];
int i,j;
for(i=0;i<16;++i)
for(j=0;j<3;++j)
scanf("%d",&a[i][j]);
while(1)
{
for(i=0;i<3;++i)
scanf("%d",&colour[i]);
if(colour[0]==-1&&colour[1]==-1&&colour[2]==-1)
break;
for(i=0;i<16;++i)
distant[i]=sqrt(pow((colour[0]-a[i][0]),2)+pow((colour[1]-a[i][1]),2)+pow((colour[2]-a[i] [2]),2));
min=distant[0];
for(i=0;i<16;++i)
if(distant[i]<=min)
min=distant[i];
for(i=0;i<16;++i)
{
if(min==distant[i])
{
printf("(%d,%d,%d) maps to (%d,%d,%d)/n",colour[0],colour[1],colour[2],a[i][0],a[i] [1],a[i][2]);
break;
}
}
}
return 0;
}