2018-2019 ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) Solution

A. Find a Number

Solved By 2017212212083

题意：$找一个最小的n使得n % d == 0 并且 n 的每一位数字加起来之和为s$

思路：

定义一个二元组$<d, s>$ 表示当前状态模d的值，以及每一位加起来的值

跑最短路，从$<0, 0>  跑到 <0, s>$

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int maxn = 1e5 + 10;
 6 const int INF = 0x3f3f3f3f;
 7 #define N 510
 8 #define M 5010
 9 
10 int d, s, cnt;
11 char ans[maxn];
12 int dis[N][M];
13 int inq[N][M];
14 int vis[N][M];
15 
16 struct node{
17     int remind;
18     int sum;
19     node(){}
20     node(int remind, int sum) :remind(remind), sum(sum){}
21 };
22 
23 void Init()
24 {
25     cnt = -1;
26     memset(inq, 0, sizeof inq);
27     memset(dis, 0x3f, sizeof dis);
28     memset(vis, 0, sizeof vis);
29 }
30 
31 void BFS()
32 {
33     queue<node>q;
34     dis[0][0] = 0;
35     q.push(node(0, 0));
36     inq[0][0] = 1;
37     while(!q.empty())
38     {
39         node st = q.front();
40         q.pop();
41         inq[st.remind][st.sum] = 0;
42         for(int i = 0; i < 10; ++i)
43         {
44             node now = node((st.remind * 10 + i) % d, st.sum + i);
45             if(now.sum > s) break;
46             if(dis[now.remind][now.sum] > dis[st.remind][st.sum] + 1)
47             {
48                 dis[now.remind][now.sum] = dis[st.remind][st.sum] + 1;
49                 if(!inq[now.remind][now.sum]) 
50                 {
51                     inq[now.remind][now.sum] = 1;
52                     q.push(now);
53                 }
54             }
55         }
56     }
57 }
58 
59 int DFS(int D,int S)
60 {
61     if(D == 0 && S == s) return 1;
62     for(int i = 0; i < 10; ++i)
63     {
64         int td = (D * 10 + i) % d;
65         int ts = S + i;
66         if(ts > s) break;
67         if(vis[td][ts]) continue;
68         if(dis[D][S] + 1 != dis[td][ts]) continue;
69         ans[++cnt] = i;
70         if(DFS(td, ts)) return 1;
71         --cnt;
72     }
73     vis[D][S] = 1;
74     return 0;
75 }
76 
77 int main()
78 {
79     while(~scanf("%d %d", &d, &s))
80     {
81         Init();
82         BFS();
83         if(dis[0][s] == INF)
84         {
85             puts("-1");
86             continue;
87         }
88         DFS(0, 0);
89         for(int i = 0; i <= cnt; ++i) printf("%d", ans[i]);
90         puts("");
91     }
92     return 0;
93 }

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