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:

Yes
2469135798

#include<iostream>
#include<string>
#include<string.h>

using namespace std;

string add(string, string);

int main(){
int ori_num[10];
int ans_num[10];
memset(ori_num, 0, sizeof(int)* 10);
memset(ans_num, 0, sizeof(int)* 10);
string ori, ans;

cin >> ori;
ans = add(ori, ori);
int length[2] = { 0, 0 };
length[0] = ori.size();
length[1] = ans.size();
for (int i = 0; i < length[0]; i++){//统计数中0~9分别出现过多少次
ori_num[ori[i] - '0'] ++;
}
for (int i = 0; i < length[1]; i++){
ans_num[ans[i] - '0'] ++;
}
for (int i = 0; i < 10; i++){
if (ans_num[i] != ori_num[i]){
cout << "No" << ans << endl;
break;
}
if (i == 9){
cout << "Yes\n" << ans <<endl;
}
}
return 0;
}

string add(string a, string b){
char ans[30];
int length = a.size();//同一个数，所以位数相同
int carry = 0;
for (int i = length-1; i >=0; i--){
ans[i] = a[i] + b[i] - '0' + carry;
if (ans[i] - '0' > 9){
ans[i] -= 10;
carry = 1;
}
else
carry = 0;
}
ans[length] = 0;
string ret = ans;
if (carry == 1)//最高位有可能有进位
ret = "1"+ret;
return ret;
}