这个数据范围让我们不得不往背包想。
但是两维相互限制。考虑坐标系旋转 \(45°\)，转化为两维不相互限制。
那我们现在相当于要安排正负号，使得 \(\sum\limits_{i = 1}^n \pm d_i\) 等于一个给定的值 \(K\)。
考虑两边加上 \(\sum\limits_{i = 1}^n d_i\) 并除以 \(2\)，转化为纯粹的 01 背包问题。
bitset 优化即可。时间复杂度 \(O(\frac{n \sum\limits_{i = 1}^n d_i}{w})\)。
code
// Problem: G - Jumping sequence
// Contest: AtCoder - AtCoder Beginner Contest 221
// URL: https://atcoder.jp/contests/abc221/tasks/abc221_g
// Memory Limit: 1024 MB
// Time Limit: 5000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 2020;
const int maxm = 3600100;
int n, A, B, a[maxn], b[maxn];
bitset<maxm> f[maxn];
void solve() {
scanf("%d%d%d", &n, &A, &B);
int s = 0;
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
s += a[i];
}
int x = A - B, y = A + B;
A = x;
B = y;
if (((A + s) & 1) || ((B + s) & 1)) {
puts("No");
return;
}
A = (A + s) / 2;
B = (B + s) / 2;
if (A < 0 || B < 0) {
puts("No");
return;
}
f[0][0] = 1;
for (int i = 1; i <= n; ++i) {
f[i] = (f[i - 1] << a[i]) | f[i - 1];
}
if (!f[n].test(A) || !f[n].test(B)) {
puts("No");
return;
}
puts("Yes");
for (int i = n; i; --i) {
if (!f[i - 1].test(A)) {
A -= a[i];
b[i] |= 2;
}
if (!f[i - 1].test(B)) {
B -= a[i];
b[i] |= 1;
}
}
for (int i = 1; i <= n; ++i) {
putchar(b[i] == 0 ? 'L' : (b[i] == 1 ? 'U' : (b[i] == 2 ? 'D' : 'R')));
}
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}