UVA12105 Bigger is Better

Bigger is Better

https://odzkskevi.qnssl.com/91b457ef612009fbd8a70f1852aad0cc?v=1508483406

【题解】

dp[i][j]表示前i位数mod m = j的最小火柴数

据此找到位数i，从高位开始枚举做即可

需要与处理iej mod k的值

mo[i][j][k]表示i * 10^j mod k的值

 

注意：：：：：：

0 mod 任何非零的数 都是 0！！！！！！！！！

 

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <vector> 
 7 #define min(a, b) ((a) < (b) ? (a) : (b))
 8 #define max(a, b) ((a) > (b) ? (a) : (b))
 9 
10 inline void swap(long long&a, long long&b)
11 {
12     long long tmp = a;a = b;b = tmp;
13 }
14 
15 inline void read(long long&x)
16 {
17     x = 0;char ch = getchar(), c = ch;
18     while(ch < '0' || ch > '9')c = ch, ch = getchar();
19     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
20 }
21 
22 const long long INF = 0x3f3f3f3f;
23 const long long MAXN = 100 + 10;
24 const long long MAXM = 3000 + 10;
25 const long long num[10] = {6,2,5,5,4,5,6,3,7,6};
26 
27 long long n,m,t,dp[MAXN][MAXM],mo[10][MAXN][MAXM];
28 
29 int  main()
30 {
31     for(register long long i = 1;i <= 9;++ i)
32         for(register long long j = 0;j <= MAXN - 10;++ j)
33             for(register long long k = 1;k <= MAXM - 10;++ k)
34                 mo[i][j][k] = (j == 0 ? i : (mo[i][j - 1][k] * 10)) % k;
35     while(scanf("%d", &n) != EOF && n)
36     {
37         ++ t;
38         printf("Case %lld: ", t);
39         read(m);
40         memset(dp, 0x3f, sizeof(dp));
41         dp[0][0] = 0;
42         for(register long long i = 0;i < n;++ i)
43             for(register long long j = 0;j < m;++ j)
44                 for(register long long k = 0;k <= 9;++ k)
45                     dp[i + 1][(j * 10 + k)%m] = min(dp[i + 1][(j * 10 + k)%m], dp[i][j] + num[k]);
46         long long flag = 0;
47         //dp[i][j]表示i位数modm = j的最小火柴数 
48         //mo[i][j][k]表示iej mod k的值 
49         for(register long long i = n;i >= 1;-- i)
50         {
51             if(dp[i][0] <= n)
52                 for(register long long k = 9;k >= 1;-- k)
53                 {
54                     long long ans = 0;
55                     if(dp[i - 1][(m - mo[k][i - 1][m])%m] + num[k] <= n)
56                     {
57                         flag = 1;
58                         printf("%lld", k);
59                         long long tmp = num[k], tmp2 =  mo[k][i - 1][m]; 
60                         for(register long long j = i - 1;j >= 1;-- j)
61                         {
62                             for(register long long kk =  9;kk >= 0;-- kk)
63                             {
64                                 if(dp[j - 1][(m - (tmp2 + mo[kk][j - 1][m]) % m) % m] + num[kk] + tmp <= n)
65                                 {
66                                     tmp2 = (tmp2 + mo[kk][j - 1][m]) % m;
67                                     tmp += num[kk];
68                                     printf("%lld", kk);
69                                     break;
70                                 }
71                             }
72                         }
73                         putchar('\n');
74                     }
75                     if(flag)break;
76                 }
77             if(flag)break;
78         }
79         if(!flag) 
80         {
81             if(n >= 6)printf("0\n");
82             else printf("-1\n");
83         }
84     } 
85     return 0;
86 }
UVA12105

 

posted @ 2017-10-25 07:51  嘒彼小星  阅读(244)  评论(0)  编辑  收藏  举报