PAT甲级练习1023. Have Fun with Numbers (20)

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

计算加倍后数字出现的频数是否和原数一致，挺简单的一题，但是也卡了一会。。。

一开始用另一个数组x2保存各数字的频数，但测试点四通不过，之后改成了直接在x1上减，判断最后是不是都为零进行判断，没想通为什么。。。

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <map>#include <set>#include <string>#include <string.h>using namespace std;char c1[22], c2[22];int x1[10],x2[10];int main(){scanf("%s", c1);int len=strlen(c1);int flag=0, tmp1, tmp2, cnt=0;for(int i=len-1; i>=0; i--){tmp1 = c1[i] - '0';x1[tmp1]++;tmp2 = 2 * tmp1 + flag;flag = tmp2 / 10;c2[cnt++] = tmp2 % 10 + '0';x1[tmp2%10]--;if(i==0 && flag==1){c2[cnt] = '1';x1[1]--;}}int flag1=0;for(int i=0; i<10; i++){if(x1[i]!=0) flag1=1;}if(flag1==0)cout<<"Yes"<<endl;elsecout<<"No"<<endl;for(int i=strlen(c2)-1; i>=0; i--){printf("%d", c2[i]-'0');}cout<<endl;cin>>c1;return 0;}