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 kk 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
说明：最开始使用的思路是大数计算的方式，存入字符串数组，然后运算后与原数组进行对比，但存在wa点，现在没找到。先放置于此：
#include <stdio.h> #include<string.h> int main() { // N为长度，M为进位 int N = 0,M = 0, i = 0, num = 0,loop = 0, j = 0; char str[100]; char tmp[100]; scanf("%s",str); N = strlen(str); for(i = N - 1; i >= 0; i--){ num = (str[i] - '0')*2 + M; M = 0; if(num > 9){ num = num % 10; M++; } tmp[i] = num + '0'; } for(i = 0; i < N; i++){ for(j = 0; j < N; j++){ if(tmp[i] == str[j]){ str[j] = 'a'; break; } } } for(i = 0; i < N; i++){ if(str[i] != 'a'){ loop = 1; break; } } if(M == 1){ printf("No 1"); for(i = 0; i < N; i++){ printf("%d",tmp[i] - '0'); } return 0; } if(loop == 1){ for(i = 0; i < N; i++){ printf("%d",tmp[i] - '0'); } }else{ printf("Yes "); for(i = 0; i < N; i++){ printf("%d",tmp[i] - '0'); } } return 0; }
后改变思路，只记录0~9十个数组出现的次数，仅需要两个长度为10的一维数组即可。使用别人写的C++代码，引用链接在最后
1 #include <cstdio> 2 #include <string> 3 #include <iostream> 4 using namespace std; 5 int cnt1[10] = { 0 }, cnt2[10] = {0}; 6 7 int main(){ 8 string s; 9 cin >> s; 10 string s2 = s; 11 for (int i = 0; i < s.size(); i++){ 12 cnt1[s[i] - '0']++; 13 } 14 int carry = 0; 15 for (int i = s.size() - 1; i >= 0; i--){ 16 s2[i] = ((s[i] - '0') * 2 + carry )% 10 + '0'; 17 carry = ((s[i] - '0') * 2 + carry) / 10; 18 } 19 if (carry){ 20 char c = carry + '0'; 21 s2 = c + s2; 22 cout << "No " << s2 << endl; 23 return 0; 24 } 25 for (int i = 0; i < s2.size(); i++){ 26 cnt2[s2[i] - '0']++; 27 } 28 bool flag = true; 29 for (int i = 0; i < 10; i++){ 30 if (cnt1[i] != cnt2[i]){ 31 flag = false; 32 break; 33 } 34 } 35 if (flag) cout << "Yes" << " " << s2 << endl; 36 else cout << "No " << s2; 37 return 0; 38 }
http://blog.csdn.net/xtzmm1215/article/details/38837203