第二题推出数学公式后发现太坑爹了。。
### oblem Statement | |||||||||||||
A *Ball Triangle* is a set of identical balls placed in a triangular shape. A Ball Triangle has **N** rows, numbered 1 to **N** from top to bottom. For all i, 1 <= i <= **N**, the i-th row contains i balls. For example, the following image shows a Ball Triangle with **N**=3.
![](https://www.ccppcoding.com/wp-content/uploads/BallTrig-1.png) Fox Jiro has infinitely many Ball Triangles. He can paint a Ball Triangle according to the following conditions: * Each of the balls has to be painted either red, green, or blue. The following image shows one valid coloring of a Ball Triangle for **N**=3. | |||||||||||||
### Definition | |||||||||||||
| |||||||||||||
### Constraints | |||||||||||||
- | **R**, **G** and **B** will each be between 0 and 1,000,000,000,000,000,000 (10^18), inclusive. | ||||||||||||
- | **N** will be between 1 and 1,000,000,000, inclusive. | ||||||||||||
### Examples | |||||||||||||
0) | |||||||||||||
| |||||||||||||
1) | |||||||||||||
| |||||||||||||
2) | |||||||||||||
| |||||||||||||
3) | |||||||||||||
| |||||||||||||
4) | |||||||||||||
| |||||||||||||
5) | |||||||||||||
| |||||||||||||
6) | |||||||||||||
|
This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.
给出N求最多可以画多少?
一个规律
给出N,那么一共有M=N*(N+1)/2个球,那么画出一个满足目标的Triangle Ball需要的球的组合设为a个R球,b个G球,c个B球。
d=M/3; a,b,c就是和d接近的三个数。
如果M%3==0,那么a=b=c等于d,这样去R,G,B中最小的一个数除a就可以得到结果了。即return min(R,G,B)/a;
如果M%3==1或==2,设这一组数为d, d, d+1或者d+1,d+1,d。
a=b=d,c=d+1或者a=b=d+1,c=d。
因为a,b总相等,那么这一组数有3中排列方式。
(a,b,c)
(a,c,b)
(c,a,b)
设第一组使用了x个,第二组使用了y个,第三组使用了z个。
所求目标是max: x+y+z
注意,我们有
xa+ya+z*c=R
xb+yc+z*a=G
xc+yb+z*b=B
左右相加,我们得到
x(a+b+c)+y(a+b+c)+z*(a+b+c)=R+G+B
整理,得到
(x+y+z)=(R+G+B)/(a+b+c)
所以x+y+z居然是固定的值!!
所以只要return (R+G+B)/(a+b+c)就可以了!
public class FoxPaintingBalls {
public long theMax(long R, long G, long B, int N){
long a, b, c;
long M=N*(N+1)/2;
long rm=M%3;
a=b=c=(M/3);
if(rm==1){
c++;
}
if(rm==2){
a++;
b++;
}
if(rm==0){
return Math.min(Math.min(R/a, G/a), B/a);
} else {
return (R+G+B)/(a+b+c);
}
}
}
如果没有注意到这个推倒而直接写一个递归函数的,就惨了,
像这样的。
private long getMax(long R, long G, long B, long a, long b, long c){
if((R==b&&G==a&&B==c)||(R==c&&G==b&&B==a)||(R==b&&G==c&&B==a)){
return 1;
}
else{
long x,y,z;
x=R-a;y=G-b;z=B-c;
long max1=0,max2=0,max3=0;
if(x>=0&&y>=0&&z>=0){
max1=getMax(x,y,z,a,b,c)+1;
}
x=R-a;y=G-c;z=B-b;
if(x>=0&&y>=0&&z>=0){
max2=getMax(x,y,z,a,b,c)+1;
}
x=R-c;y=G-a;z=B-b;
if(x>=0&&y>=0&&z>=0){
max3=getMax(x,y,z,a,b,c)+1;
}
return Math.max(Math.max(max1, max2), max3);
}
}
原文链接: https://www.cnblogs.com/gaoqichao/archive/2012/08/17/2643811.html
欢迎关注
微信关注下方公众号,第一时间获取干货硬货;公众号内回复【pdf】免费获取数百本计算机经典书籍
原创文章受到原创版权保护。转载请注明出处:https://www.ccppcoding.com/archives/59868
非原创文章文中已经注明原地址,如有侵权,联系删除
关注公众号【高性能架构探索】,第一时间获取最新文章
转载文章受原作者版权保护。转载请注明原作者出处!