[PAT (Advanced Level) ]1023. Have Fun with Numbers 解题文档

1023. Have Fun with Numbers (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

分析：此题不能使用int能求解，明显越界。考虑到long也不是很好，因此采用string来存储数字。用vector来标记数量（也可以用静态数组，一样的）

难度不大。

C++代码如下。

#include<iostream>
#include<vector>
#include<string>
#include<strstream>
using namespace std;
vector<int> vec1,vec2;
string doubleResult;
string doubleNum(string str){
string res="";
int N=(int)str.size();
int xxx=0;
for(int i=0;i<N;i++){
int t=(str[N-i-1])-48;
vec1[t]++;
int temp=((t*2)%10+xxx);
vec2[temp]++;
strstream ss;
string strtemp;
ss<<temp;
ss>>strtemp;
res=strtemp+res;
if(t*2<10){
xxx=0;
}
if(t*2>=10){
xxx=1;
}
}
if(xxx==1){
res="1"+res;
}
return res;
}

string isFun(string str){
for(int i=0;i<10;i++){//initial
vec1.push_back(0);
vec2.push_back(0);
}
doubleResult=doubleNum(str);
for(int i=0;i<10;i++){
if(vec1[i]!=vec2[i])
return "No";
}
return "Yes";
}

int main(){
string str;
cin>>str;
cout<<isFun(str)<<endl<<doubleResult;
}

测试点均通过。