CodeForces1070A Find a Number 图论

令状态$f(i, j)$表示模$d$为$i$，和为$j$时的最小数

可以通过$bfs$来转移

然而就没了...

复杂度$O(10ds)$

#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
    #define ri register int
    #define ll long long
    #define pii pair<int, int>
    #define mp make_pair
    #define pb push_back
    #define fi first
    #define se second
    #define tpr template <typename ra>
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)
    #define gc getchar
    inline int read() {
        int p = , w = ; char c = gc();
        while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
        while(c >= '' && c <= '') p = p *  + c - '', c = gc();
        return p * w;
    }
}
using namespace std;
using namespace remoon;
#define sid 525
#define pid 5205

int d, s;
struct node {
    int d, s, di;
    node() {}
    node(int a, int b, int c) : d(a), s(b), di(c) {}
};
queue <pii> q;

bool vis[sid][pid];
node pre[sid][pid];

void bfs() {
    vis[][] = ;
    q.push(mp(, ));
    while(!q.empty()) {
        pii p = q.front(); q.pop();
        rep(i, , ) {
            int nd = (p.fi *  + i) % d;
            int ns = p.se + i;
            if(ns > s) break;
            if(!vis[nd][ns]) {
                vis[nd][ns] = ;
                q.push(mp(nd, ns));
                pre[nd][ns] = node(p.fi, p.se, i + '');
            }
        }
    }
}

inline void dfs(int nd, int ns) {
    if(pre[nd][ns].di != )
    dfs(pre[nd][ns].d, pre[nd][ns].s);
    if(pre[nd][ns].di != )
    printf("%c", pre[nd][ns].di);
}

int main() {
    d = read(); s = read();
    bfs();
    if(!vis[][s]) puts("-1");
    else dfs(, s);
    return ;
}