【HDU2089】不要62【数位DP】【记忆化搜索】

【题目链接】

忘记写了个判断条件

如果上一位为6，当前为也为6，那么状态还是1。

/* Telekinetic Forest Guard */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 20;

int dig[maxn];
LL dp[maxn][3];

// 0 ÎÞ 62
// 1 ÓÐ 6
// 2 ÓÐ 62

template <class numtype>
inline void read(numtype &x) {
int f = 0; x = 0; char ch = getchar();
for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? 1 : 0;
for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
if(f) x = -x;
}

inline LL dfs(int pos, int state, bool limit) {
if(pos == 0) return state != 2;
if(!limit && ~dp[pos][state]) return dp[pos][state];

int upb = limit ? dig[pos] : 9;
LL res = 0;
for(int i = 0; i <= upb; i++) if(i ^ 4) {
int s = state;
if(state == 1) {
if(i == 2) s = 2;
else if(i != 6) s = 0;
}
else if(state == 0 && i == 6) s = 1;
res += dfs(pos - 1, s, limit && i == dig[pos]);
}

if(!limit) dp[pos][state] = res;
return res;
}

inline LL calc(LL x) {
int top = 0;
for(; x; x /= 10) dig[++top] = x % 10;
return dfs(top, 0, 1);
}

int main() {
LL l, r;
memset(dp, -1, sizeof(dp));
while(1) {
read(l); read(r);
if(!l && !r) break;
printf("%lld\n", calc(r) - calc(l - 1));
}
return 0;
}