rt，思路是把影响横纵坐标的元素拆成两个数组然后折半搜索看能不能凑出横纵坐标值。

测评记录：https://atcoder.jp/contests/abc326/submissions/47053248

#include <bits/stdc++.h>

#define int long long
#define N 1100

using namespace std;

inline int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while(c < '0' || c > '9'){if(c == '-') f = -1; c = getchar();}
    while(c <= '9' && c >= '0') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x * f;
}

int n, X, Y, a[N], a1[N], a2[N], m1, m2, now1, now2, A1, A2;
unordered_map<int, int> mp;

inline void dfs1(int x, int sum)
{
    // cout << "sum1 : " << sum << endl;
    if(x == m1 / 2 + 1) return mp[sum] = (A1 | (1ll << 50)), void();
    A1 |= (1 << x);
    dfs1(x + 1, sum + a1[x]);
    A1 ^= (1 << x);
    dfs1(x + 1, sum - a1[x]);
    return ;
}

inline void dfs2(int x, int sum)
{
    // cout << "sum2  : " << sum << endl;
    if(x == m1 + 1)
    {
        if(mp[Y - sum]) now1 = (A2 | mp[Y - sum]);
        // cout << " S SD: " << Y - sum << " : " << mp[Y - sum];
        return ;
    }
    A2 |= (1 << x);
    dfs2(x + 1, sum + a1[x]);
    if(now1) return ;
    A2 ^= (1 << x);
    dfs2(x + 1, sum - a1[x]);
    return ;
}

inline void dfs3(int x, int sum)
{
    if(x == m2 / 2 + 1) return mp[sum] = (A1 | (1ll << 50)), void();
    A1 |= (1 << x);
    dfs3(x + 1, sum + a2[x]);
    A1 ^= (1 << x);
    dfs3(x + 1, sum - a2[x]);
    return ;
}

inline void dfs4(int x, int sum)
{
    if(x == m2 + 1)
    {
        if(mp[X - sum]) now2 = (A2 | mp[X - sum]);
        return ;
    }
    A2 |= (1 << x);
    dfs4(x + 1, sum + a2[x]);
    if(now2) return ;
    A2 ^= (1 << x);
    dfs4(x + 1, sum - a2[x]);
    return ;
}

inline void print(int x)
{
    while(x)
    {
        cout << ((x & 1) ? 1 : 0);
        x /= 2;
    }
    cout << endl;
    return ;
}

signed main()
{
    n = read(), X = read(), Y = read();
    for(int i = 1; i <= n; i ++)
    {
        a[i] = read();
        if(i & 1) a1[++ m1] = a[i];
        else a2[++ m2] = a[i];
    }
    // for(int i = 1; i <= m1; i ++) cout << a1[i] << " "; cout << endl;
    // for(int i = 1; i <= m2; i ++) cout << a2[i] << " "; cout << endl;
    dfs1(1, 0);
    dfs2(m1 / 2 + 1, 0);
    mp.clear();
    A1 = A2 = 0;
    if(m2 != 0) dfs3(1, 0);
    // cout << "cao" << endl;
    dfs4(m2 / 2 + 1, 0);
    // print(now1);
    // print(now2);
    if(now1 == 0 || now2 == 0) return cout << "No" << endl, 0;
    cout << "Yes" << endl;
    int t = 1, ss1 = 0, ss2 = 0;
    for(int i = 1; i <= n; i ++)
    {
        if(i & 1)
        {
            int k = (i % 2 == 0 ? i / 2 : i / 2 + 1);
            if(t == 1)
            {
                if(now1 & (1 << k)) t = 2, cout << "L";
                else t = 4, cout << "R";
            }
            else if(t == 3)
            {
                if(now1 & (1 << k)) t = 2, cout << "R";
                else t = 4, cout << "L";
            }
            if(now1 & (1 << k)) ss1 += a1[k];
            else ss1 -= a1[k];
        }
        else
        {
            int k = (i % 2 == 0 ? i / 2 : i / 2 + 1);
            if(t == 2)
            {
                if(now2 & (1 << k)) t = 1, cout << "R";
                else t = 3, cout << "L";
            }
            if(t == 4)
            {
                if(now2 & (1 << k)) t = 1, cout << "L";
                else t = 3, cout << "R";
            }
            if(now2 & (1 << k)) ss2 += a2[k];
            else ss2 -= a2[k];
        }
    }
    // cout << endl;
    // cout << ss1 << endl;
    // cout << ss2 << endl;
    return 0;
}
/*
6 0 9
1 2 3 4 5 6
*/