PAT (Advanced Level) 1023. Have Fun with Numbers (20) 字符串翻倍，哈希

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different
permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.
Sample Input:

1234567899

Sample Output:

Yes
2469135798

字符串处理。使用一个数组a[10]作哈希表判断是否匹配。

/*2015.7.22cyq*/
#include <iostream>
#include <vector>
#include <string>
using namespace std;

string doubleString(const string &s1){
string s2(s1);
int carry=0;
int tmp;
for(auto it=s2.rbegin();it!=s2.rend();it++){
tmp=(*it-'0')*2+carry;
carry=tmp/10;
*it=tmp%10+'0';
}
if(carry>0)
s2.insert(s2.begin(),1,carry+'0');
return s2;
}
int main(){
string s1;
cin>>s1;
string s2=doubleString(s1);
vector<int> a(10,0);
int n1=s1.size();
int n2=s2.size();
if(n2>n1)
cout<<"No"<<endl;
else{
bool match=true;
for(int i=0;i<n1;i++){
a[s1[i]-'0']++;
}
for(int i=0;i<n2;i++){
if(--a[s2[i]-'0']<0){
match=false;
break;
}
}
if(match)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
cout<<s2<<endl;
return 0;
}

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