这一题要采用逆向思维,不要去判断每个数字出现的有没有这么多,而是去统计数字出现的次数,反过来构造出对应的描述数字,如果描述数字跟原来的数字相等,那就是self described的数
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
string self_describe(string lineBuffer) {
string res = "";
size_t size = lineBuffer.size();
int b[10] = {0};
for(size_t i=0; i<size; i++) {
++b[lineBuffer[i]-'0'];
}
for(size_t i=0; i<size; i++) {
res += (b[i]+'0');
}
return res;
}
int main (int argc, char* argv[])
{
ifstream file;
string lineBuffer;
file.open(argv[1]);
while (!file.eof()) {
getline(file, lineBuffer);
if (lineBuffer.length() == 0)
continue; //ignore all empty lines
else {
if(lineBuffer == self_describe(lineBuffer))
cout << "1\n";
else
cout << "0\n";
}
}
return 0;
}