SGU Self-number2

我们考虑，因为最多n为10^7, 7*9 = 63，也就是说一个数x如果不是self-number，那么他的原数一定在[x-63, x]之间。

由于空间有限，所以我们可以用大小>=64的数组优化（只要能存下前63个数即可）。

另外一个优化，打表处理10^4之内的每位数字之和， 则d(x) = x + sum[x/10000] + sum[x%10000];

还有一个坑点是， 查询数字的值不一定是递增的， 也可能有重复的查询。。。

 

 1 /*Author :usedrose  */
 2 /*Created Time :2015/7/24 11:09:55*/
 3 /*File Name :2.cpp*/
 4 #include <cstdio>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <sstream>
 8 #include <cstdlib>
 9 #include <cstring>
10 #include <climits>
11 #include <vector>
12 #include <string>
13 #include <ctime>
14 #include <cmath>
15 #include <deque>
16 #include <queue>
17 #include <stack>
18 #include <set>
19 #include <map>
20 #define INF 0x3f3f3f3f
21 #define eps 1e-8
22 #define pi acos(-1.0)
23 #define MAXN 10110
24 #define OK cout << "ok" << endl;
25 #define o(a) cout << #a << " = " << a << endl
26 #define o1(a,b) cout << #a << " = " << a << "  " << #b << " = " << b << endl
27 using namespace std;
28 typedef long long LL;
29 
30 struct node {
31     int num, index, ans;
32 }a[MAXN];
33 bool f[100];
34 int n, m;
35 int sum[MAXN];
36 
37 bool cmp1(node a, node b)
38 {
39     return a.index < b.index;
40 }
41 
42 bool cmp2(node a, node b)
43 {
44     return a.num < b. num;
45 }
46 void init()
47 {
48     for (int i = 1;i <= 10000; ++ i)
49     {
50         int k = i;
51         while (k) {
52             sum[i] += k%10;
53             k /= 10;
54         }
55     }
56 }
57 
58 int main()
59 {
60     //freopen("data.in","r",stdin);
61     //freopen("data.out","w",stdout);
62     cin.tie(0);
63     ios::sync_with_stdio(false);
64     init();
65 
66     while (cin >> n >> m) {
67         for (int i = 1;i <= m; ++ i) {
68             cin >> a[i].index;
69             a[i].num = i;
70         }
71         sort(a+1, a+1+m, cmp1);
72         int di = 0, cur = 1;
73         memset(f, 1, sizeof(f));
74         for (int i = 1;i <= n; ++ i) {
75             if (f[i%64]) {
76                 di++;
77                 while (a[cur].index == di && cur <= m) {
78                     a[cur].ans = i;
79                     cur++;
80                 }
81             }
82             f[i%64] = 1;
83             int nxt = i + sum[i/10000] + sum[i%10000];
84             f[nxt%64] = 0;
85         }
86 
87         cout << di << endl;
88         sort(a+1, a+1+m, cmp2);
89         cout << a[1].ans;
90         for (int i = 2;i <= m; ++ i)
91             cout << " " << a[i].ans; 
92         cout << endl;
93     }
94     return 0;
95 }
View Code

 

posted @ 2015-07-24 12:44  UsedRose  阅读(99)  评论(0)  编辑  收藏  举报