BZOJ1833 [ZJOI2010]count 数字计数

裸的数位DP我还不会= =b

令f[i][j][k]表示长度为i的数，开头为j，数字k的个数，这个可以预处理出来

只要计算[0, b] - [0, a - 1]即可，稍微讨论一下什么的就好了

/**************************************************************
    Problem: 1833
    User: rausen
    Language: C++
    Result: Accepted
    Time:28 ms
    Memory:828 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
 
using namespace std;
typedef long long ll;
 
struct data {
  ll a[10];
  inline ll& operator [] (int x) {
    return a[x];
  }
  inline data operator + (data x) {
    data t;
    int i;
    for (i = 0; i < 10; ++i)
      t.a[i] = a[i] + x.a[i];
    return t;
  }
  inline data operator += (data x) {
    *this = *this + x;
    return *this;
  }
} p1, p2, f[25][10];
 
ll a, b, t[25];
 
void pre_work() {
  int i, x, y;
  for (t[1] = 1, i = 2; i <= 15; ++i)
    t[i] = t[i - 1] * 10;
  for (i = 0; i <= 9; ++i)
    f[1][i][i] = 1;
  for (i = 2; i <= 12; ++i)
    for (x = 0; x <= 9; ++x)
      for (y = 0; y <= 9; ++y) {
    f[i][y] += f[i - 1][x];
    f[i][y][y] += t[i - 1];
      }
}
 
void work(data &p, ll x) {
  int len = 15, i, j, now;
  if (x == 0) {
    p[0] = 1;
    return;
  }
  while (t[len] > x) --len;
  for (i = 1; i < len; ++i)
    for (j = 1; j <= 9; ++j)
      p += f[i][j];
  ++p[0];
  now = x / t[len];
  for (i = 1; i < now; ++i)
    p += f[len][i];
  x %= t[len];
  p[now] += x + 1;
  for (i = len - 1; i; --i) {
    now =  x / t[i];
    for (j = 0; j < now; ++j)
      p += f[i][j];
    x %= t[i];
    p[now] += x + 1;
  }
}
 
int main() {
  int i;
  pre_work();
  scanf("%lld%lld", &a, &b);
  work(p1, a - 1);
  work(p2, b);
  for (i = 0; i < 9; ++i)
    printf("%lld ", p2[i] - p1[i]);
  printf("%lld", p2[9] - p1[9]);
  return 0;
}