标签:
给定矩形的长和宽,判断它们能否构成长方体的六个面
1 #include<iostream> 2 #include<string> 3 using namespace std; 4 int t; 5 void change(int a[7][3]) //输入两个数字,按顺序:小,大;排列。 6 { 7 for (int k = 1; k <= 6; k++) 8 if (a[k][1] > a[k][2]) 9 { 10 t = a[k][1]; 11 a[k][1] = a[k][2]; 12 a[k][2] = t; 13 } 14 } 15 int main() 16 { 17 int a[7][3]; 18 for (int i = 1; i <= 6; i++) 19 for (int ii = 1; ii <= 2; ii++) 20 cin >> a[i][ii]; 21 change(a); 22 int coincide[7] = { 0 }; 23 int square = 0; 24 int side[7][3] = { 0 }; 25 for (int k = 1; k <= 6; k++) 26 { 27 for (int kk = 1; kk <= 6; kk++) 28 { 29 if (a[k][1] == a[kk][1] && a[k][2] == a[kk][2] && k != kk) //两面完全相等 30 coincide[k]++; 31 for (int j = 1; j <= 2; j++) 32 { 33 if (a[k][1] == a[kk][j] && k != kk) 34 side[k][1]++; 35 if (a[k][2] == a[kk][j] && k != kk) 36 side[k][2]++; 37 } 38 } 39 if (a[k][1] == a[k][2]) 40 square++; 41 } 42 int judge = 1; // 43 if (square == 0 && coincide[0] == 5) //// 44 judge = 0; 45 if (square > 2 && side[0][0] != 10) //square大于2即为正方形。出去自身所在边,side应为10 46 judge = 0; 47 for (int h = 1; h <= 6; h++) 48 { 49 if (coincide[h] == 0) //没有面面相同 50 51 { 52 judge = 0; break; 53 } 54 55 for (int j = 1; j <= 2; j++) 56 if (side[h][j] < 3) //自己算一条,总共四条,应该为3。 57 { 58 judge = 0; 59 break; 60 } 61 } 62 if (judge == 0) cout << "sorry!wrong!"; 63 else cout << "possible"; 64 system("pause"); 65 return 0; 66 }
标签:
原文地址:http://www.cnblogs.com/ghost-song/p/4448835.html