2 条题解

  • 1
    @ 2025-10-26 19:21:56

    #include<bits/stdc++.h> using namespace std; int m,n; string minn="999999999999999999999999999999999999999999999999999999999999999999",s; char a[40][40]; bool t(string x,string y){ if(x.size()<y.size())return 1; if(x.size()>y.size())return 0; for(int i=0;i<x.size();i++){ if(int(x[i]-'0')<int(y[i]-'0'))return 1; if(int(x[i]-'0')>int(y[i]-'0'))return 0; } return 0; } int main(){ cin>>n>>m;

    for(int i=1;i<=n;i++){
    	for(int j=1;j<=m;j++){
    		cin>>a[i][j];
    	}
    }
    for(int i=1;i<=n;i++){
    	for(int j=1;j<=m;j++){
    		s="";
    		a[i][m+1]='#';
    		a[i][0]='*';
    		bool f=0;
    		if(a[i][j]!='#'&&(a[i][j-1]<'0'||a[i][j-1]>'9')){
    			f=1;
    			for(int k=j;k<=m;k++){
    				if(a[i][k]=='#')break;
    				s+=a[i][k];
    			}
    		}
    		if(s.size()>0)
    		if(t(s,minn))minn=s;
    	}
    }
    cout<<minn;
    return 0;
    

    }

    • 0
      @ 2022-5-21 20:52:05

      #include<bits/stdc++.h> using namespace std; long long n,m,mink=25; char ch; string s,minn="zzzzzzzzzzzzzzzz"; int main() { cin>>n>>m; for(int i=1;i<=n;i++) { s=""; for(int j=1;j<=m;j++) { cin>>ch; if(ch=='#') { if(s!=""&&(s.size()==mink&&s<minn||s.size()<mink)) { mink=s.size(); minn=s; } s="";//空字符串 } else s+=ch; }//一次判断字符串 if(s!=""&&(s.size()==mink&&s<minn||s.size()<mink)) { mink=s.size(); minn=s; }//二次判断字符串 } cout<<minn; return 0; }

      • 1

      信息

      ID
      90
      时间
      1000ms
      内存
      256MiB
      难度
      6
      标签
      (无)
      递交数
      65
      已通过
      22
      上传者