https://ac.nowcoder.com/acm/contest/330/B
C++版本一
题解:
std
这道题的做法比较开放。只要按题意构造就好了。
大体思路参照下图。
根据 n 和 m 的奇偶分类讨论。
特殊的边界数据:1行2列或2行1列的情况。
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
if (n == 1)
{
if (m == 2)
cout << "RL" << endl;
else
cout << -1 << endl;
}
else if (m == 1)
{
if (n == 2)
cout << "DU" << endl;
else
cout << -1 << endl;
}
else if (n % 2 == 0)
{
cout << "R";
for (int i = 0; i < n; i += 2)
{
if (i)
cout << "D";
for (int j = 1; j < m - 1; j++)
cout << "R";
cout << "D";
for (int j = 1; j < m - 1; j++)
cout << "L";
}
cout << "L";
for (int i = 0; i < n - 1; i++)
cout << "U";
cout << endl;
}
else if (m % 2 == 0)
{
cout << "D";
for (int i = 0; i < m; i += 2)
{
if (i)
cout << "R";
for (int j = 1; j < n - 1; j++)
cout << "D";
cout << "R";
for (int j = 1; j < n - 1; j++)
cout << "U";
}
cout << "U";
for (int i = 0; i < m - 1; i++)
cout << "L";
cout << endl;
}
else
cout << -1 << endl;
return 0;
}
’C++版本二
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
int ans,cnt,flag,temp;
int a[N][N];
string str;
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
scanf("%d%d",&n,&m);
//scanf("%d",&t);
//while(t--){}
if(m==2&&n==1){
cout<<"RL"<<endl;
return 0;
}
if(n==2&&m==1){
cout<<"DU"<<endl;
return 0;
}
if((n%2&&m%2)||n==1||m==1){
cout<<-1<<endl;
}else if(n%2==0){
for(int i=1;i<n;i++)
str+='D';
str+='R';
for(int i=1;i<=n/2;i++){
for(int j=1;j<m-1;j++)
str+='R';
str+='U';
for(int j=1;j<m-1;j++)
str+='L';
if(i!=n/2)
str+='U';
}
str+='L';
cout<<str<<endl;
}else if(m%2==0){
for(int i=1;i<m;i++)
str+='R';
str+='D';
for(int i=1;i<=m/2;i++){
for(int j=1;j<n-1;j++)
str+='D';
str+='L';
for(int j=1;j<n-1;j++)
str+='U';
if(i!=m/2)
str+='L';
}
str+='U';
cout<<str<<endl;
}
//cout << "Hello world!" << endl;
return 0;
}