POJ 1016 循环数 字符串处理与英文理解

    这题挺简单，但是题目稍微有点长，基本没什么算法。

#include <iostream>#include <string>using namespace std;string transfer(string s){int num[10] = {0};char res[80];int i;int j = 0;for (i = 0 ; i < s.length() ; i++)num[s[i] - '0']++;for ( i = 0 ; i < 10; i++){if (num[i] > 0 && num[i] < 10){res[j++] = num[i] + '0';res[j++] = i + '0';}else if(num[i] >= 10){res[j++] = num[i]/10 + '0';res[j++] = num[i]%10 + '0';res[j++] = i + '0';}}res[j++] = '/0';return res;} int main(){string s[20];int i,j;bool flag;//标记是否输出过while (1){for (i = 0;i < 20; i++){s[i] = "";}flag = false;cin>>s[0];if(s[0] == "-1") break;for( i = 0;i < 16; i++){s[i+1] = transfer(s[i]);}for (i = 0;i < 16 && (!flag); i++){for (j = i+1;j<16 && (!flag);j++){if (s[j] == s[i] && j == 1 && i == 0){cout<<s[0]<<" is "<<"self-inventorying"<<endl;flag = true;break;}else if (s[j] == s[i] && j == i+1 && i != 0){cout<<s[0]<<" is "<<"self-inventorying after "<<j-1<<" steps"<<endl;flag = true;break;}else if (s[j] == s[i] && j != i+1){cout<<s[0]<<" enters an inventory loop of length "<<(j - i)<<endl;flag = true;break;}}}if (!flag){cout<<s[0]<<" can not be classified after 15 iterations"<<endl;}}return 0;}