1023. Have Fun with Numbers (20)
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:1234567899Sample Output:
Yes 2469135798
点击打开链接Pat甲 1023
题意就是看这个数字乘以二后，和原来数字的组成是不是相同，用两个数组标记下比一下就行了
坑点是如果组成不同。也需要打印数字的两倍。
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int num[30],res[30],count1[10],count2[10];
char str1[30];
int main()
{scanf("%s",&str1);int len=strlen(str1);int index=29;for(int i=len-1;i>=0;i--){num[index--]=str1[i]-'0';count1[str1[i]-'0']++;}int carry=0;for(int i=29;i>=0;i--){res[i]=(carry+num[i]*2)%10;carry=(carry+num[i]*2)/10;}index=0;while(res[index]==0)index++;for(int i=index;i<=29;i++)count2[res[i]]++;for(int i=0;i<=9;i++){if(count1[i]!=count2[i]){printf("No\n");for(int k=index;k<=29;k++){printf("%d",res[k]);} printf("\n");return 0;}}printf("Yes\n");for(int i=index;i<=29;i++){printf("%d",res[i]);}printf("\n");return 0;
}