poj 1016 Numbers That Count【字符串】

题目大意：给你一个数字串，最大长度80，然后计算里面每个数字出现的次数，按照从小到大的顺序排列成另一个数字串。比如5553141变化后是21 13 14 35（2个1,1个3,1个4,3个5）。

 

如果1次变化后，数字串没变，那么输出“n is self-inventorying”其中，n代表题目给你的那个字符串。比如31123314 按照规则变化后还是31123314，所以输出“31123314 is self-inventorying ”。

 

如果 J 次变化后，字符串变成了一个self-inventorying 那么输出“n is self-inventorying after j steps”其中，n代表题目给你的那个字符串，j 就是变化次数了；比如给的是 21221314 第一次变化得到 31321314 ，再次变化得到 31123314，而31123314再次变化还是31123314，所以他是一个self-inventorying，所以变化了2次，输出 “21221314 is self-inventorying after 2 steps ”。

 

如果J次变化后，字符串串和前面某个字符串相同，也就是说形成了一个环，则输出环的长度。例如 314213241519 变化后是412223241519，再次变化是 314213241519，和第一个字符串重复了，所以环的长度是2。输出“314213241519 enters an inventory loop of length 2”

 

如果15次变化后还是没有出现上面的情况，输出“n can not be classified after 15 iterations”，n代表原串。

#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<map>
#include<stdlib.h>
using namespace std;

string str[20];
map<char, int> mp;
map<char, int>::iterator it;

void change(int i)
{
    str[i+1].erase();
    mp.clear();
    int len = str[i].length();

    for(int j = 0; j < len; j++) {
        mp[ str[i][j] ]++;
    }

    for(it = mp.begin(); it != mp.end(); it++) {
        char s[10];
     //      itoa(it->second, s, 10);
        sprintf(s, "%d", it->second);
        str[i+1] += s;
        str[i+1] += it->first;
    }
}

int main()
{
    while(cin >> str[0]) {
        if(str[0] == "-1") break;
        change( 0 );
        if(str[1] == str[0]) {
            cout << str[0] << " is self-inventorying\n";
            continue;
        }
        int step = 1;
        bool bl = 0;
        int tot;
        while(1) {
            change( step );
            if(str[step] == str[step-1]) {
                bl = 0;
                break;
            }
            if(!bl) {
                for(int i = 0; i < step; i++) {
                    if(str[step] == str[i]) {
                        tot = step - i;
                        bl = 1;
                        break;
                    }
                }
            }
            step++;
            if(step > 15) break;
        }
        if(step > 15 && !bl) {
            cout << str[0] << " can not be classified after 15 iterations\n";
        } else if( bl ) {
            cout << str[0] << " enters an inventory loop of length " << tot <<endl;
        } else {
            cout << str[0] << " is self-inventorying after " << step-1 << " steps\n";
        }
    }
    return 0;
}

测试数据:

输入:

22
31123314
314213241519
21221314
111222234459
123456789
654641656
2101400052100005496
10000000002000000000
333
1
99999999999999999999999999999999999999999999999999999999999999999999999999999999
0000
0001
0111
1111
123456789
456137892
123213241561
543265544536464364
5412314454766464364
543267685643564364
5423434560121016464364
-1

输出 :

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
123456789 is self-inventorying after 5 steps
654641656 enters an inventory loop of length 2
2101400052100005496 enters an inventory loop of length 2
10000000002000000000 enters an inventory loop of length 3
333 is self-inventorying after 11 steps
1 is self-inventorying after 12 steps
99999999999999999999999999999999999999999999999999999999999999999999999999999999 can not be classified after 15 iterations
0000 enters an inventory loop of length 2
0001 is self-inventorying after 8 steps
0111 is self-inventorying after 8 steps
1111 is self-inventorying after 8 steps
123456789 is self-inventorying after 5 steps
456137892 is self-inventorying after 5 steps
123213241561 enters an inventory loop of length 2
543265544536464364 enters an inventory loop of length 2
5412314454766464364 is self-inventorying after 3 steps
543267685643564364 enters an inventory loop of length 2
5423434560121016464364 is self-inventorying after 3 steps