数字三角
一个无限等边三角形网格(数字表示网点的序号)，如下面图所示：
1
/ \
2---3
/ \ / \
4---5---6
/ \ / \ / \
7---8---9--10
/ \ / \ / \ / \
11--12--13--14--15
/ \ / \ / \ / \ / \
16--17--18--19--20--21
...........................
例如：
{1，2，3}、{3，8，10}、{1，11，15}是等边三角形的顶点集合
{1，2，3，4}和{3，8，10，14}是平行四边形的顶点集合
{2，3，4，6，8，9}是六边形的顶点集合
写一个算法，判断输入的集合属于可以接受的图形
一个图形要成为可接受的图形，必须满足如下两个条件：
1、图形每条边必须和网格中的某条边重合(如：{2，9}是重合的，而{2，13}则不是)；
2、图形每条边必须相等。
输入
元素个数为3、4、6的元素列表
输出
是否可接受
输入出范例
1,2,3=true
1,2,3,4=false
2,3,4,6,8,9=true
原帖地址http://community.csdn.net/Expert/topic/5369/5369826.xml?temp=.7778742
下面是我给出的答案：
1
#include "stdafx.h"
2#include <list>
3#include <iostream>
4#include "stdio.h"
5using namespace std;
6
7bool CalAngle(int *num, int *lines, int agrc);
8int GetLine(int exp);
9
10int _tmain(int argc, _TCHAR* argv[])
11
{
12
if (argc <= 1)
13
{
14 return 0;
15
}
16
17 int *num = new int[argc - 1];
18 int *lines = new int[argc - 1];
19
20
21 if (num == NULL || lines == NULL)
22 {
23 cout << "分配内存失败！" << endl;
24 return 0;
25 }
26
27 memset(num, 0, argc - 1);
28 memset(lines, 0, argc - 1);
29
30 for(int i = 1; i < argc; i++)
31 {
32 sscanf((char*)argv[i], "%d", num + i - 1);
33 }
34 if (CalAngle(num, lines, argc - 1))
35 {
36 cout << "true" << endl;
37 }
38 else
39 {
40 cout << "false" << endl;
41 }
42 delete[] num;
43 delete[] lines;
44 return 0;
45
}
46
47//求两点之间距离
48int GetDistance(int m, int n, int l1, int l2)
49{
50 int temp = n;
51 if (m < n)
52 {
53 temp = m;
54 m = n;
55 n = temp;
56 }
57
58 if (l1 < l2)
59 {
60 temp = l1;
61 l1 = l2;
62 l2 = temp;
63 }
64
65 //判断是否在同一层上
66 int ilen = 0;
67 if (l1 == l2 )
68 {
69 //ic++;
70 ilen = m - n; //abs(m - n);
71 return ilen;
72 }
73
74 //左边
75 int imax = m;
76 int icurvalue = n;
77 for(int k = l2; k < l1; k++)
78 {
79 icurvalue += k;
80 if ( icurvalue > imax)
81 {
82 break;
83 }
84 else if (icurvalue == imax)
85 {
86 ilen = abs(l1 - l2);
87 return ilen;
88 }
89 }
90
91 //右边
92 icurvalue = n;
93 for(int i = l2; ilen == 0 && i < l1; i++)
94 {
95 icurvalue += (i + 1);
96 if ( icurvalue > imax)
97 {
98 break;
99 }
100 else if (icurvalue == imax)
101 {
102 ilen = l1 - l2;
103 return ilen;
104 }
105 }
106 return 0;
107}
108
109bool CalAngle(int *num, int *lines, int points)
110{
111 int ilen = 0;
112 int ioldlen = 99999;
113 int *pd = new int[points * points]; //各个点之间距离
114 memset(pd, 0, points * points);
115
116 for(int i = 0; i < points; i++)
117 {
118 *(lines + i) = GetLine(*(num + i));
119 }
120
121 //求出两点之间的最小距离
122 for(int i = 0; i < points; i++)
123 {
124 for( int j = 0; j < points; j++)
125 {
126 if (i == j)
127 continue;
128 ilen = GetDistance(num[i], num[j], lines[i], lines[j]);
129 pd[i * points + j] = ilen;
130 if (ilen > 0 && ilen < ioldlen)
131 {
132 ioldlen = ilen;
133 }
134 }
135 }
136
137 //某一点到相邻的点的距离等于最小距离的个数不为2就不符合条件
138 for(int i = 0; i < points; i++)
139 {
140 int ic = 0;
141 for( int j = 0; j < points; j++)
142 {
143 if (i == j)
144 {
145 continue;
146 }
147 //int ilen = GetDistance(num[i], num[j], lines[i], lines[j]);
148 if ( pd[i * points + j] == ioldlen )
149 {
150 ic++;
151 }
152 }
153
154 if (ic != 2)
155 {
156 delete[] pd;
157 return false;
158 }
159 }
160
161 delete[] pd;
162 return true;
163}
164
165//获取某个数字所在的行数
166int GetLine(int exp)
167{
168 long sum = 0;
169 for(int i = 1; i < 10000; i++)
170 {
171 sum += i;
172 if ( exp <= sum )
173 {
174 return i;
175 }
176 }
177}
178
#include "stdafx.h"2
#include <list> 3
#include <iostream>4
#include "stdio.h"5
using namespace std; 6
7
bool CalAngle(int *num, int *lines, int agrc);8
int GetLine(int exp);9
10
int _tmain(int argc, _TCHAR* argv[])11
{
12
if (argc <= 1)13
{
14
return 0;15
}16
17
int *num = new int[argc - 1];18
int *lines = new int[argc - 1];19
20
21
if (num == NULL || lines == NULL)22
{23
cout << "分配内存失败！" << endl;24
return 0;25
}26
27
memset(num, 0, argc - 1);28
memset(lines, 0, argc - 1);29
30
for(int i = 1; i < argc; i++)31
{32
sscanf((char*)argv[i], "%d", num + i - 1);33
}34
if (CalAngle(num, lines, argc - 1))35
{36
cout << "true" << endl;37
}38
else39
{40
cout << "false" << endl;41
}42
delete[] num;43
delete[] lines;44
return 0;45
}46
47
//求两点之间距离48
int GetDistance(int m, int n, int l1, int l2)49
{50
int temp = n;51
if (m < n)52
{53
temp = m;54
m = n;55
n = temp;56
}57
58
if (l1 < l2)59
{60
temp = l1;61
l1 = l2;62
l2 = temp;63
}64
65
//判断是否在同一层上66
int ilen = 0;67
if (l1 == l2 )68
{69
//ic++;70
ilen = m - n; //abs(m - n);71
return ilen;72
}73
74
//左边75
int imax = m;76
int icurvalue = n;77
for(int k = l2; k < l1; k++)78
{79
icurvalue += k;80
if ( icurvalue > imax)81
{82
break;83
}84
else if (icurvalue == imax)85
{86
ilen = abs(l1 - l2);87
return ilen;88
}89
}90
91
//右边92
icurvalue = n;93
for(int i = l2; ilen == 0 && i < l1; i++)94
{95
icurvalue += (i + 1);96
if ( icurvalue > imax)97
{98
break;99
}100
else if (icurvalue == imax)101
{102
ilen = l1 - l2;103
return ilen;104
}105
}106
return 0;107
}108
109
bool CalAngle(int *num, int *lines, int points)110
{111
int ilen = 0;112
int ioldlen = 99999;113
int *pd = new int[points * points]; //各个点之间距离114
memset(pd, 0, points * points);115
116
for(int i = 0; i < points; i++)117
{118
*(lines + i) = GetLine(*(num + i));119
}120
121
//求出两点之间的最小距离122
for(int i = 0; i < points; i++)123
{124
for( int j = 0; j < points; j++)125
{126
if (i == j)127
continue;128
ilen = GetDistance(num[i], num[j], lines[i], lines[j]);129
pd[i * points + j] = ilen;130
if (ilen > 0 && ilen < ioldlen)131
{132
ioldlen = ilen;133
}134
}135
}136
137
//某一点到相邻的点的距离等于最小距离的个数不为2就不符合条件138
for(int i = 0; i < points; i++)139
{140
int ic = 0;141
for( int j = 0; j < points; j++)142
{143
if (i == j)144
{145
continue;146
}147
//int ilen = GetDistance(num[i], num[j], lines[i], lines[j]);148
if ( pd[i * points + j] == ioldlen )149
{150
ic++;151
}152
}153
154
if (ic != 2)155
{156
delete[] pd;157
return false;158
}159
}160
161
delete[] pd;162
return true;163
}164
165
//获取某个数字所在的行数166
int GetLine(int exp)167
{168
long sum = 0;169
for(int i = 1; i < 10000; i++)170
{171
sum += i;172
if ( exp <= sum )173
{174
return i;175
}176
}177
}178
