【PAT】1023. Have Fun with Numbers (20)
2013-08-21 16:52
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题目:http://pat.zju.edu.cn/contests/pat-a-practise/1023
分析:简单题。用俩个数组标记数字0~9出现的数次就可以了。
题目描述:
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:
Sample Output:
分析:简单题。用俩个数组标记数字0~9出现的数次就可以了。
题目描述:
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.h> using namespace std; #define max 22 char input[max]; int flag_1[11]; int flag_2[11]; int s[max]; bool isSame = true; int main() { int len,i,j,temp; int jinwei; //进位 //清零 memset(flag_1,0,sizeof(flag_1)); memset(flag_2,0,sizeof(flag_2)); cin>>input; len = strlen(input); for(j=0,i=len-1; i>=0; i--,j++) { s[j] = input[i] - '0'; flag_1[ s[j] ] ++; } jinwei = 0; //进位 //乘以2 for(i=0; i<len; i++) { temp = ( s[i]*2 + jinwei )/10; s[i] = (s[i]*2 + jinwei)%10; flag_2[ s[i] ] ++; jinwei = temp; } for(j=1; j<=10; j++) { if(flag_2[j] != flag_1[j]) { isSame = false; break; } } if(isSame) cout<<"Yes"<<endl; else cout<<"No"<<endl; //输出 if(jinwei > 0) { j=i; s[i] = jinwei;} //最高位有进位 else j = --i; for(; j>=0; j--) { cout<<s[j]; } cout<<endl; return 0; }
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