PAT 甲级 1023 Have Fun with Numbers (20 分)（permutation是全排列题目没读懂）

1023 Have Fun with Numbers (20 分)

 

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 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

解题思路：
　　给定一个数，乘以2之后，各个数位的出现次数，是否与乘之前是否相同，相同输出Yes，否认输出No，然后下一行打印出乘2之后的各个数位。
　　这个题目是属于超大数运算的类型，题目给出的最高数位是20位，那么即便是用最高的unsigned long long 也会面临溢出的情况，所以输入和输出，只能用string，诸位乘2，然后再记录每一位出现的次数，相比较就行。

     大数乘法！全排列的意思是：各个数字出现的次数分别相同！

AC代码：

#include<bits/stdc++.h>
using namespace std;
char a[21];
char b[21];
int ori[11];
int ne[11];
int main(){
    memset(ori,0,sizeof(ori));
    memset(ne,0,sizeof(ne));
    cin>>a;
    //先反着放
    int l=strlen(a);
    for(int i=l-1;i>=0;i--){
        b[l-i]=a[i];
        ori[a[i]-'0']++;
    } 
    //大数乘法 
    int x=0;
    for(int i=1;i<=l;i++){
        int y=b[i]-'0';
        y*=2;y+=x;
        b[i]=y%10+'0';
        x=y/10;    
        ne[b[i]-'0']++;            
    }
    int l2=l;
    if(x!=0){
        l2++;
        b[l2]=x+'0';
        ne[b[l2]-'0']++;    
    }
    //判断 
    int f=1;
    for(int i=0;i<=9;i++){
        if(ne[i]!=ori[i]){
            f=0;
            break;
        }
    }
    if(f){
        cout<<"Yes"<<endl;
    }else{
        cout<<"No"<<endl;
    }
    for(int i=l2;i>=1;i--){
        cout<<b[i];
    }
    return 0;
}