POJ1016 Numbers That Count

题目来源:http://poj.org/problem?id=1016

题目大意:

  对一个非负整数定义一种运算(inventory):数这个数中各个数字出现的次数,然后按顺序记录下来。比如“5553141”有2个1,1个3,一个4,3个5,于是运算后为“21131435”,对于这种运算有的数字有着有趣的性质:如“31123314”,它运算后的结果和它本身是一样的(self-inventorying)。如果做j次inventory运算后,第j次迭代的结果是self-inventorying的,则称这个数是j步后self-inventory.此外,如果n进行j次迭代后,再进行k(k>=2)次迭代,结果与第j次迭代的结果相等,称它 enters an inventory loop of length k.

  写一个程序,判断一系列非负整数属于上面三种数字的哪一种(只考虑15次迭代)。

输入:一系列的非负整数,每个最多80位,-1标志结束。

输出:对于每一个输入的整数n,根据该数的性质输出一下四条语句中的一条:

n is self-inventorying 
n is self-inventorying after j steps 
n enters an inventory loop of length k 
n can not be classified after 15 iterations


Sample Input

22 
31123314 
314213241519 
21221314 
111222234459 
-1

Sample Output

22 is self-inventorying 
31123314 is self-inventorying 
314213241519 enters an inventory loop of length 2 
21221314 is self-inventorying after 2 steps 
111222234459 enters an inventory loop of length 2

这道题我就是用简单的模拟做的,具体实现见代码。

 1 //////////////////////////////////////////////////////////////////////////
 2 //        POJ1016 Numbers That Count
 3 //        Memory: 272K        Time: 47MS
 4 //        Language: C++        Result: Accepted
 5 //////////////////////////////////////////////////////////////////////////
 6 
 7 #include <iostream>
 8 #include <string>
 9 
10 using namespace std;
11 
12 string int2str(int n) {
13     string str;
14     if (n == 0)
15         str = "0";
16     while (n != 0) {
17         str.insert(str.begin(), '0' + n % 10);
18         n /= 10;
19     }
20     return str;
21 }
22 
23 int main() {
24     while (true) {
25         string strList[16];
26         int digCount[10];
27         int j = 0, k = 0;
28         cin >> strList[0];
29         if (strList[0][0] == '-') {
30             break;
31         }
32         int count = 0;
33         bool flag = false;
34         for (int i = 1; i < 16; ++i) {
35             for (int k = 0; k < 10; ++k) {
36                 digCount[k] = 0;
37             }
38             for (int t = 0; t < strList[i - 1].size(); ++t) {
39                 int d = strList[i - 1][t] - '0';
40                 ++digCount[d];
41             }
42             strList[i] = "";
43             for (int k = 0; k < 10; ++k) {
44                 if (digCount[k] == 0) {
45                     continue;
46                 }
47                 strList[i] += int2str(digCount[k]);
48                 strList[i] += int2str(k);
49             }
50             if (strList[i] == strList[i - 1]) {
51                 if (i == 1) {
52                     cout << strList[0] << " is self-inventorying" << endl;
53                     flag = true;
54                     break;
55                 } else {
56                     cout << strList[0] << " is self-inventorying after " << i - 1 <<" steps" << endl;
57                     flag = true;
58                     break;
59                 }
60             }
61             for (int j = 0; j < i; ++j) {
62                 if (strList[i] == strList[j]) {
63                     cout << strList[0] << " enters an inventory loop of length " << i - j << endl;
64                     flag = true;
65                     break;
66                 }
67             }
68             if (flag == true) {
69                 break;
70             }
71         }
72         if (flag == true) {
73             continue;
74         } else {
75             cout << strList[0] << " can not be classified after 15 iterations" << endl;
76         }
77     }
78     system("pause");
79 }
View Code
posted @ 2013-07-31 14:44  小菜刷题史  阅读(219)  评论(0编辑  收藏  举报