[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:

Yes2469135798

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。