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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
#pragma warning(disable:4996)#include <stdio.h>#include <iostream>#include <string>using namespace std;int main(void) {string s1 = "12345678901234567890",s2="12345678901234567890";cin >> s1;int n = s1.length();int jinwei = 0;for (int i = 0; i < n; i++) {if(s1[i]<‘5‘)s2[i] = 2 * (s1[i] - ‘0‘)+‘0‘;else {if (i > 0) {s2[i] = 2 * (s1[i] - ‘0‘) - 10+‘0‘;s2[i - 1] += 1;}if (i == 0) {s2[i] = 2 * (s1[i] - ‘0‘) - 10+‘0‘;jinwei = 1;}}}if (jinwei == 1) {cout << "No" << endl;cout << 1;for (int i = 0; i < n; i++) {cout << s2[i];}return 0;}int a[20] = { 0 };for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (s2[i] == s1[j] && a[j] == 0)a[j]++;}}int count = 0;for (int i = 0; i < n; i++)if (a[i] == 1)count++;if (count == n) {cout << "Yes"<<endl;for (int i = 0; i < n; i++) {cout << s2[i];}}else {cout << "No"<<endl;for (int i = 0; i < n; i++) {cout << s2[i];}}return 0;}
1023. Have Fun with Numbers (20)
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原文地址:http://www.cnblogs.com/zzandliz/p/5023083.html