D. Pawn

// Problem: D. Pawn
// Contest: Codeforces - Codeforces Beta Round #40 (Div. 2)
// URL: https://codeforces.com/problemset/problem/41/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 2022-03-04 17:50:29
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
const int N = 1e2 + 10;
int n, m, p;
int a[N][N];
int f[N][N][20];
vector<char> path;
int s;
void dfs (int x, int y, int cnt) {
    if (x == n) {
        s = y;
        return ;
    }
    // cout << x << " " << y << endl;
    for (int r = 0; r < p; r ++) {
        if (y != 1)
        if (f[x + 1][y - 1][r] + a[x][y] == f[x][y][cnt]) {
            path.pb('R');
            dfs(x + 1, y - 1, r);
            break;
        }
        if (y != m) {
            if (f[x + 1][y + 1][r] + a[x][y]== f[x][y][cnt]) {
                path.pb('L');
                dfs(x + 1, y + 1, r);
                break;
            }
        }
    }
}
void solve() {
    cin >> n >> m >> p;
    p ++;
    for (int i = 1; i<= n ;i ++) {
        string s;
        cin >> s;
        for (int j =1;j <= m; j ++)
            {
                a[i][j] = s[j - 1] - '0';
            }
    }
    memset(f, -0x3f, sizeof f);
    for (int j = 1;j <= m; j ++)
        {
            f[n][j][a[n][j] % p]= a[n][j];
        }
        
    for (int i = n;  i>= 2;  i--) {
        for (int j = 1; j <= m; j ++) {
            for (int k = 0; k < p; k ++) {               
                int x;
                if (j !=  1)
                {
                    x = a[i - 1][j - 1];
                    f[i - 1][j - 1][(k +x) % p] = max (f[i - 1][j - 1][(k + x) % p], f[i][j][k] + x);
                }
                if (j != m) {
                    x = a[i - 1][j + 1];
                    f[i - 1][j + 1][(k +x) % p] = max (f[i - 1][j + 1][(k + x) % p], f[i][j][k] + x);
                }
            }
        }
    }
    int ans = -1, id;
    for (int i = 1; i <= m; i ++)
        {
            if (ans < f[1][i][0]) {
                ans = f[1][i][0];
                id = i;
            }
        }

    cout << ans << endl;
    if (ans == -1) {
        return;
    }
    dfs(1, id, 0);
    cout << s << endl;
    for (int i = path.size() - 1; i >= 0; i --)
        cout << path[i];
    
}
int main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

> 设置dp状态的时候不能解决，就加一维，直到解决

这里通过增加一维r然后转移很好设计。正向转移