PAT:A1023 Have Fun with Numbers
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
代码:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct bign{
int d[30];
int len;
bign() {
len = 0;
memset(d, 0, sizeof(d));
}
};
// 2. change
bign change(char a[]) {
bign c;
c.len = strlen(a);
for(int i = 0; i < c.len; i++) {
c.d[i] = a[c.len - 1 - i] - '0';
}
return c;
}
// 3. compare
int compare(bign a, bign b) {
if(a.len > b.len) return 1;
else if(a.len < b.len) return -1;
else {
for(int i = 0; i < a.len; i++) {
if(a.d[i] > b.d[i]) return 1;
else if(a.d[i] < b.d[i]) return -1;
}
return 0;
}
}
// 4. print
void print(bign a) {
for(int i = a.len - 1; i >= 0; i--) {
printf("%d", a.d[i]);
}
}
// 5. 高精度整数 与 低精度整数的乘法
bign mult(bign a, int b) {
bign c;
int carry = 0;
for(int i = 0; i < a.len; i++) {
int temp = a.d[i] * b + carry;
c.d[c.len++] = temp % 10;
carry = temp / 10;
}
while(carry != 0) {
c.d[c.len++] = carry % 10;
carry = carry / 10;
}
return c;
}
// 6.cmp
bool cmp(int a, int b) {
return a < b;
}
int main() {
char A[30];
scanf("%s", A);
bign a = change(A);
bign c = mult(a, 2);
bign d = c;
sort(a.d, a.d+a.len);
sort(c.d, c.d+c.len);
int x = compare(a, c);
if(x == 0) printf("Yes\n");
else printf("No\n");
print(d);
return 0;
}