1 /****************************************************
 2 PAT B1027 打印沙漏
 3 ****************************************************/
 4 #include <iostream>
 5 #include <cmath>
 6 
 7 using namespace std;
 8 
 9 int main() {
10     char symbol;
11     int store = 0;
12 
13     cin >> store >> symbol;
14 
15     int maxEdgeSize = int(sqrt(2.0 * (store + 1))) - 1;                  //最大边的符号个数
16     maxEdgeSize = maxEdgeSize % 2 == 0 ? maxEdgeSize - 1 : maxEdgeSize;  //最大边的符号个数肯定是奇数
17 
18     int usedCont = 0;                       //使用的符号个数
19     int triLevel = (maxEdgeSize - 1) / 2;   //沙漏上部三角形的层数
20     //输出沙漏上不部分的倒三角
21     for (int i = 0; i <= triLevel; ++i) {
22         for (int j = 0; j < i; ++j) cout << ‘ ‘;
23         for (int j = 0; j < maxEdgeSize - 2 * i; ++j) {
24             cout << symbol;
25             usedCont++;
26         }
27         cout << endl;
28     }
29 
30     //输出沙漏剩余的部分
31     for (int i = 1; i <= triLevel; ++i) {
32         for (int j = 0; j < triLevel - i; ++j) cout << ‘ ‘;
33         for (int j = 0; j < 1 + 2 * i; ++j) {
34             cout << symbol;
35             usedCont++;
36         }
37         cout << endl;
38     }
39 
40     //输出剩余符号个数
41     cout << store - usedCont;
42 
43     return 0;
44 }

1 //获取1100以内所有能组成的沙漏包含的符号数
2 int cost = 1;
3 
4 cout << cost << ", ";
5 for (int i = 1; cost < 1100; ++i) {
6     cost += 2 * (2 * i + 1);
7 
8     cout << cost << ", ";
9 }

 1 /****************************************************
 2 PAT B1027 打印沙漏
 3 ****************************************************/
 4 #include <iostream>
 5 
 6 using namespace std;
 7 
 8 //1100个字符时所有可能组成的沙漏的消耗
 9 const int COST[23] = { 1, 7, 17, 31, 49, 71, 97, 127, 161, 199, 241, 287, 337, 391, 449, 511, 577, 647, 721, 799, 881, 967, 1057 };
10 
11 //获取最大可输出漏斗的下三角的层数
12 int myFind(int left, int right, int num) {
13     int mid = 0;
14     while (left < right) {
15         mid = (left + right) / 2;
16         if (right - left == 1) {
17             return left;
18         }
19         else if (COST[mid] == num) {
20             return mid;
21         }
22         else {
23             COST[mid] > num ? right = mid : left = mid;
24         }
25     }
26 }
27 
28 int main() {
29     char symbol;
30     int store = 0;
31 
32     cin >> store >> symbol;
33 
34     //获取最大可输出漏斗的下三角的层数
35     int triLevel = myFind(0, 22, store);
36 
37     //输出漏斗上面的部分
38     for (int i = 0; i < triLevel; ++i) {
39         for (int j = 0; j < i; ++j) cout << ‘ ‘;
40         for (int j = 0; j < 2 * (triLevel - i) + 1; ++j) cout << symbol;
41         cout << endl;
42     }
43 
44     //输出漏斗下面的三角形
45     for (int i = 0; i <= triLevel; ++i) {
46         for (int j = 0; j < triLevel - i; ++j) cout << ‘ ‘;
47         for (int j = 0; j < 2 * i + 1; ++j) cout << symbol;
48         cout << endl;
49     }
50 
51     //输出剩余符号数
52     cout << store - COST[triLevel];
53     
54     return 0;
55 }