hdu 2089 不要62(数位dp)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <set>
#include <list>
#include <map>
#include <limits>
using namespace std;
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))
#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))
#define MAXN 1000000
/***************BEGINHERE*****************/
int f[15][10], a[15];
int n, m;

int Count(int x) {
int tmp = x, cnt = 0;
if (!x) return 0;
// 得到各个数位 从cnt-1至0是从高到低
while (tmp) {
a[cnt++] = tmp%10;
tmp /= 10;
}
int ret = 0;
a[cnt] = -1;
for (int k=cnt-1;k>=0;--k) {
for (int i=a[k]-1;i>=0;--i) {
if (a[k+1]==6&&i==2) continue;
ret += f[k+1][i];
}
if (a[k+1]==6&&a[k]==2) break;
if (a[k] == 4) break;
}
//cout << ret << endl;
return ret;
}

int main() {
freopen("input.in", "r", stdin);
REP(i, 0, 9)
f[0][i] = 0;
f[0][0] = 1;
REP(k, 1, 7)
REP(i, 0, 9) {
f[k][i] = 0;
REP(j, 0, 9) {
if (i == 4) continue;
if (i == 6 && j == 2) continue;
f[k][i] = f[k][i] + f[k-1][j];
}
}
while(scanf("%d%d",&n,&m) && (n|m)) {
cout << Count(m+1) - Count(n) << endl;
}
return 0;
}