C - Back and Forth (atcoder.jp)

输出从(sx,sy) 到(tx,ty)的四条路径，每条线路之间不能经过同一个点（除终点和原点外） 

#include <iostream>
#include <vector>
#include <bits/stdc++.h>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <set>
#include <stack>
#include <algorithm>
#define x first
#define y second
#define pb emplace_back
#define fu(i,a,b) for(int i=a;i<=b; ++ i)
#define fd(i,a,b) for(int i=a;i>=b;	-- i)
#define endl '
'
#define ms(x,y) memset(x,y,sizeof x)
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef vector<vector<LL>> VVL;
typedef vector<vector<int>> VVI;
typedef vector<LL> VL;
typedef vector<bool> VB;
typedef vector<int> VI;
typedef vector<string> VS;
typedef pair<int,int> PII;
typedef vector<PII> VPII;
typedef pair<PII,int> PIII;
typedef pair<double,double> PDD;
typedef pair<double,int> PDI;
typedef pair<char,int> PCI;
typedef pair<string,int> PSI;
typedef pair<int,string> PIS;
typedef pair<int,char> PIC;
typedef pair<LL,LL> PLL;
typedef __int128 i128;
typedef unsigned long long ULL;
const int N =1e2 + 10 ,M = N * 2 ,INF = 0x3f3f3f3f,P = 131;
const double eps = 1e-8,DNF = 1e18;
const int mod = 998244353 ,base= 20010;
const LL LNF=(LL) INF * INF;



void solve()
{
	int sx,sy,tx,ty;cin >> sx >> sy >> tx >> ty;
	
	int dx = tx - sx,dy = ty - sy;
	
	cout << string(dy,'U') << string(dx,'R') ;
	cout << string(dy,'D') << string(dx,'L') ;
	cout <<'L' <<  string(dy + 1 ,'U') << string(dx + 1 ,'R') << 'D';
	cout <<'R' <<  string(dy + 1 ,'D') << string(dx + 1 ,'L') << 'U';
	
	
	cout << endl;
}

signed main()
{
//  freopen("1.txt","r",stdin);
//	#define int long long 
//	init(N-1);
    ios
//	cout << fixed<<setprecision(2);
    int t=1;
//   	cin>>t;
    int now = 1;
    while(t -- )
    {
//      cout<<"Case "; 
//      cout<<"Case #"; 
//      cout<<"Scenario #"; 
//      cout<< now ++ <<": ";
//      cout<< now ++ <<": 
";
		solve();
    }


    return 0;
}