PAT甲级—1023 Have Fun with Numbers (20 分)
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 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<cstring>
using namespace std;
int main(){
long long a,sum;
while(scanf("%lld",&a)!=EOF){
sum=a*2;
// printf("%lld",sum);
string b=to_string(a);
int a1[25]={
0};
int b1[25]={
0};
for(int i=0;i<b.length();i++){
a1[b[i]-'0']++;
}
string s=to_string(sum);
for(int i=0;i<s.length();i++){
b1[s[i]-'0']++;
}
int flag=0;
for(int i=0;i<s.length();i++){
if(a1[i]!=b1[i]){
printf("No\n");
flag=1;
}
}
if(!flag){
printf("Yes\n");
}
printf("%lld",sum);
}
return 0;
}
部分错误–9分
参考答案
#include <cstdio>
#include <string.h>
using namespace std;
int book[10];
int main() {
char num[22];
scanf("%s", num);
int flag = 0, len = strlen(num);
for(int i = len - 1; i >= 0; i--) {
int temp = num[i] - '0';
book[temp]++;
temp = temp * 2 + flag;
flag = 0;
if(temp >= 10) {
temp = temp - 10;
flag = 1;
}
num[i] = (temp + '0');
book[temp]--;
}
int flag1 = 0;
for(int i = 0; i < 10; i++) {
if(book[i] != 0)
flag1 = 1;
}
printf("%s", (flag == 1 || flag1 == 1) ? "No\n" : "Yes\n");
if(flag == 1) printf("1");
printf("%s", num);
return 0;
}