标签：div nta style hat main exce def cin pie
You are given two positive integer numbers a and b. Permute (change order) of the digits of a to construct maximal number not exceeding b. No number in input and/or output can start with the digit 0.
It is allowed to leave a as it is.
The first line contains integer a (1 ≤ a ≤ 1018). The second line contains integer b (1 ≤ b ≤ 1018). Numbers don‘t have leading zeroes. It is guaranteed that answer exists.
Print the maximum possible number that is a permutation of digits of a and is not greater than b. The answer can‘t have any leading zeroes. It is guaranteed that the answer exists.
The number in the output should have exactly the same length as number a. It should be a permutation of digits of a.
123
222
213
3921
10000
9321
4940
5000
4940
一道cf的c题,挺好的,直接用字符串的操作可以很快的解决.
1 #include <iostream> 2 #include <algorithm> 3 #define ll long long int 4 using namespace std; 5 string s,ss; 6 string st; 7 int main(){ 8 cin>>s>>ss; 9 sort(s.begin(),s.end()); 10 if(ss.length()>s.length()){ 11 reverse(s.begin(),s.end()); 12 cout<<s<<endl; 13 return 0; 14 } 15 for(int i=0;i<s.length();i++) 16 for(int j=s.length()-1;j>i;j--){ 17 st=s; 18 swap(s[i],s[j]); 19 sort(s.begin()+i+1,s.end()); 20 if(s.compare(ss)>0) 21 s=st; 22 else break; 23 } 24 cout<<s<<endl; 25 return 0; 26 }
标签：div nta style hat main exce def cin pie
原文地址：https://www.cnblogs.com/zllwxm123/p/9314589.html
