SGU - 355 - Numbers Painting （贪心）
355. Numbers Painting
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
Memory limit: 65536 kilobytes
input: standard
output: standard
output: standard
Dr. Vasechkin wants to paint all numbers from 1 to N in such a way that if number A is divisible by number B, numbers A and B have different colors.
Help Dr. Vasechkin to find such a painting, where the number of the colors used is minimal.
Input
The input contains integer number
N (
).
Output
Write the number of the colors
M in the desired painting in the first line of the output. In the second line of the output write the desired painting of numbers from 1 to
N. The used colors should be represented by numbers from 1 to
M. If there are several solutions, choose any of them.
Example(s)
sample input |
sample output |
12 |
4 1 2 2 3 2 3 2 4 3 3 2 4 |
| Online Contester Team © 2002 - 2010. All rights reserved. |
思路：给1到N的数字涂颜色，使得颜色数尽可能少，贪心.....
首先可以看出1可以整除任何数，即任何数都不可以与1相同颜色，然后可以发现互质的一群数可以相同颜色，而当两个数有大于1的约数时可以判断这两个数不同颜色，慢慢可以看出1,2,4,8,16,32......这些数的颜色都不一样，到32为止有6种颜色，可以验证是最小的颜色种类使用数，于是归纳出解决方案为一个筛法（与素数筛类似，具体看代码）
注意打印数字颜色的方法，对于数字i，从i~n*i(n*i为小于N的最大数),对于每个区间（m， m*2-1）取一种颜色（m为i的倍数）.......（自己想想为什么，有别的好的想法可以讨论）
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int color[10005];
int main() {
int N;
while(scanf("%d", &N) != EOF) {
int ans = 1;
for(int i = 2; i <= N / 2 + 1; i++) { //算出最小的颜色种类数
int tmp = 1;
for(int j = i; j <= N; j *= 2) {
tmp++;
}
if(tmp > ans) ans = tmp;
}
memset(color, 0, sizeof(color)); //记录每个数字的颜色
color[1] = 1;
for(int i = 2; i <= N; i++) {
if(color[i] == 0) {
int c = 2;
int t = i;
for(int j = i; j <= N; j += i) {
if(j >= t * 2) c++, t *= 2;
color[j] = c;
}
}
}
printf("%d\n", ans); //输出结果
for(int i = 1; i < N; i++) printf("%d ", color[i]);
printf("%d\n", color[N]);
}
return 0;
}