1023 Have Fun with Numbers （20 分)
标签：which original put result div another str pen tst
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.
Each input contains one test case. Each case contains one positive integer with no more than 20 digits.
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.
1234567899
Yes
2469135798
#include<bits/stdc++.h> using namespace std; typedef long long ll; #define maxnum 100005 int a[30] = {0}; int count1[10] = {0}; //分别对倍乘前后的位计数 int count2[10] = {0}; int main(){ string s; cin >> s; int len = s.size(); for(int i=0;i < s.size();i++){ a[i] = s[i]-‘0‘; } for(int i=0;i < len;i++){ count1[a[i]]++; } for(int i=0;i < len/2;i++){ //翻转 swap(a[i],a[len-i-1]); } int jinwei = 0; //倍乘 for(int i=0;i <= len;i++){ int num = a[i]*2 + jinwei; a[i] = num%10; jinwei = (num)/10; } // for(auto num:a) cout << num << " "; int pos = 0; for(int i=30;i >= 0;i--){ if(a[i]){pos = i;break;} } for(int i=0;i <= pos;i++){ count2[a[i]]++; } int flag = 1; for(int i=0;i < 10;i++){ if(count1[i] != count2[i])flag = 0; } if(flag) cout << "Yes" << endl; else cout << "No" << endl; for(int i=pos;i >= 0;i--){ cout << a[i]; } return 0; }
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PAT 1023 Have Fun with Numbers
标签：which original put result div another str pen tst
原文地址：https://www.cnblogs.com/cunyusup/p/10696692.html
