1 solutions
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0
C++ :
#include<iostream> #include<cstdio> using namespace std; int main() { string n; int s; cin>>n; cin>>s; int l=n.size(); for(int i=1;i<=s;++i) { for(int j=0;j<l-1;++j) if(n[j]>n[j+1]) { for(int k=j;k<l-1;k++) n[k]=n[k+1]; break; } l--;; } bool f=0; for(int i=0;i<=l-1;i++) { if(n[i]!='0') f=true; if(f)cout<<n[i]; } }
Pascal :
var s:string;min:array[1..240]of char;p:array[0..240]of integer;n,i,j,l:integer; begin readln(s); l:=length(s); readln(n); n:=l-n; fillchar(min,sizeof(min),58); for i:=1 to n do for j:=p[i-1]+1 to l-n+i do if s[j]<min[i] then begin p[i]:=j;min[i]:=s[j] end; i:=1; while min[i]=#48 do inc(i); while i<=n do begin write(min[i]);inc(i) end; writeln end.
Python :
# coding=utf-8 a=list(map(lambda x: int(x), input().split()))[0] s=[] while a!=0: s.append(a%10) a = a // 10 s.reverse() n=list(map(lambda x: int(x), input().split()))[0] s1=s for i in range(n): l=len(s) f=1 for j in range(l-1): if s[j]>s[j+1]: s.remove(s[j]) f=0 break if f==1 and j==l-2: s.remove(s[l-1]) #print(s) s2=[] for i in s: s2.append(i) for i in s: if i==0: #print(i) s2.remove(i) else: break for i in s2: print (i,end='')
- 1
Information
- ID
- 703
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By