2 条题解
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1
#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;}
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0
#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