POJ 3286 How many 0's? （数位dp）

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#define MOD 1000000007
#define INF 0x7fffffff
using namespace std;
typedef long long ll;
int bit[70];
ll f[70][20][5];

ll dp(int pos, int sta, int flag, int str)
{
//printf("%d %d %d %d\n", pos, sta, flag, str);
//第len到pos+1位已确定，现在要确定第pos位
//前面已经出现了sta个c
//flag=1，不论第pos位为什么都比给定的数小
//str=1，第len到第pos+1位出现了不是0的数
if(pos == 0) return sta;
if(flag && f[pos][sta][str] != -1) return f[pos][sta][str];
ll ans = 0;
int x = flag ? 9 : bit[pos];
for(int i = 0; i <= x; i++)
{
if(i == 0)
{
if(str) ans += dp(pos - 1, sta + 1, flag || i < x, 1);
else ans += dp(pos - 1, sta, flag || i < x, 0);
}
else
{
ans += dp(pos - 1, sta, flag || i < x, 1);
}
}
if(flag) f[pos][sta][str] = ans;
return ans;
}

ll cal(ll x)
{
if(x == -1) return -1;
int len = 0;
while(x)
{
bit[++len] = x % 10;
x /= 10;
}
ll ans = dp(len, 0, 0, 0);
//printf("---%I64d---\n", ans);
return ans;
}

int main()
{
#ifdef LOCAL
freopen("dpdata.txt", "r", stdin);
#endif

ll m, n;
memset(f, -1, sizeof(f));
while(scanf("%I64d %I64d", &m, &n) != EOF)
{
if(m < 0 && n < 0) break;
//printf("m=%I64d n=%I64d\n", m, n);
printf("%I64d\n", cal(n) - cal(m - 1));
}

return 0;
}