计算直线的交点数
Time Limit:1000MS Memory Limit:65536K
Total Submit:260 Accepted:119
Description
平面上有n条直线,且无三线共点,问这些直线能有多少种不同交点数。
比如,如果n=2,则可能的交点数量为0(平行)或者1(不平行)。
Input
输入数据包含多个测试实例,每个测试实例占一行,每行包含一个正整数n(n <= 20),n表示直线的数量.
Output
每个测试实例对应一行输出,从小到大列出所有相交方案,其中每个数为可能的交点数,每行的整数之间用一个空格隔开。
Sample Input
2
3
Sample Output
0 1
0 2 3
Source
hdu1466
题目意思很明确,网上有很多的解释都差不多,我记得从前我做这个题目的时候感到很纠结,主要还是看代码的时候不容易看懂,现在稍微好一点了,总之一定要自己去画,这样才会理解!!
网上代码如下(我再解释一下):
#include<stdio.h>
#include<string.h>
int main()
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{
int i,j,n,f[21][191]; //f[i][j]代表i条直线是否能产生j个交点,如果能f[i][j]=1,否则为0;
memset(f,0,sizeof(f));
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for(i=0;i<21;i++) //零个交点置零
f[i][0]=1;
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for(n=2;n<21;n++) //从两条直线开始求
for(i=n-1;i>=1;i--) //取出n-i条做变化
for(j=0;j<191;j++) //j变化
if(f[n-i][j]==1) //如果取出的n-i条能过产生j个交点,置f[n][j+(n-i)*i]=1,j+(n-i)*i为取出n-i条直线做变化情况下的交点数
f[n][j+(n-i)*i]=1;
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while(scanf("%d",&n)!=EOF)
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{
printf("0");
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for(j=1;j<=n*(n-1)/2;j++) //统计
if(f[n][j])
printf(" %d",j);
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printf("\n");
}
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return 0;
}
其实可以直接这么写,容易理解一点。。
代码如下:
#include<string.h>
#include<stdio.h>
int main()
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{
int i,j,n;
int f[21][191];
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memset(f,0,sizeof(f));
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for(i=1;i<21;i++)
f[i][0]=1;
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for(j=1;j<191;j++)
f[1][j]=0;
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for(n=2;n<21;n++)
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{
for(i=1;i<n;i++)
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{
for(j=0;j<191;j++)
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{
if(f[i][j])
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{
f[n][j+i*(n-i)]=1;
}
}
}
}
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while(scanf("%d",&n)!=EOF)
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{
printf("0");
for(j=1;j<=n*(n-1)/2;j++)
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{
if(f[n][j])
printf(" %d",j);
}
printf("\n");
}
return 0;
}
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posted on 2010-09-14 10:46
jince 阅读(939)
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