BZOJ1833 [ZJOI2010]count 数字计数

 1 /**************************************************************
 2     Problem: 1833
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:28 ms
 7     Memory:828 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cstring>
12  
13 using namespace std;
14 typedef long long ll;
15  
16 struct data {
17   ll a[10];
18   inline ll& operator [] (int x) {
19     return a[x];
20   }
21   inline data operator + (data x) {
22     data t;
23     int i;
24     for (i = 0; i < 10; ++i)
25       t.a[i] = a[i] + x.a[i];
26     return t;
27   }
28   inline data operator += (data x) {
29     *this = *this + x;
30     return *this;
31   }
32 } p1, p2, f[25][10];
33  
34 ll a, b, t[25];
35  
36 void pre_work() {
37   int i, x, y;
38   for (t[1] = 1, i = 2; i <= 15; ++i)
39     t[i] = t[i - 1] * 10;
40   for (i = 0; i <= 9; ++i)
41     f[1][i][i] = 1;
42   for (i = 2; i <= 12; ++i)
43     for (x = 0; x <= 9; ++x)
44       for (y = 0; y <= 9; ++y) {
45     f[i][y] += f[i - 1][x];
46     f[i][y][y] += t[i - 1];
47       }
48 }
49  
50 void work(data &p, ll x) {
51   int len = 15, i, j, now;
52   if (x == 0) {
53     p[0] = 1;
54     return;
55   }
56   while (t[len] > x) --len;
57   for (i = 1; i < len; ++i)
58     for (j = 1; j <= 9; ++j)
59       p += f[i][j];
60   ++p[0];
61   now = x / t[len];
62   for (i = 1; i < now; ++i)
63     p += f[len][i];
64   x %= t[len];
65   p[now] += x + 1;
66   for (i = len - 1; i; --i) {
67     now =  x / t[i];
68     for (j = 0; j < now; ++j)
69       p += f[i][j];
70     x %= t[i];
71     p[now] += x + 1;
72   }
73 }
74  
75 int main() {
76   int i;
77   pre_work();
78   scanf("%lld%lld", &a, &b);
79   work(p1, a - 1);
80   work(p2, b);
81   for (i = 0; i < 9; ++i)
82     printf("%lld ", p2[i] - p1[i]);
83   printf("%lld", p2[9] - p1[9]);
84   return 0;
85 }