hdoj 2089 不要62

题目链接：不要62

题目大意：给你一个区间，问这个区间里面符合条件的个数有多少，条件是不能出现62连号或者出现4

题目思路：直接数位dp就好，还不需要判断前导0的情况

#include <map>#include <set>#include <queue>#include <stack>#include <cmath>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;int a[20],dp[20][2];int dfs(int pos,int pre,int sta,bool limit){    if(pos == -1) return 1;    if(!limit&&dp[pos][sta] != -1) return dp[pos][sta];    int up = limit?a[pos]:9;    int tmp = 0;    for(int i = 0;i <= up;i++){        if(pre == 6&&i == 2||i == 4) continue;        tmp += dfs(pos-1,i,i == 6,limit&&i == a[pos]);    }    if(!limit) dp[pos][sta] = tmp;    return tmp;}int solve(int x){    int pos = 0;    while(x){        a[pos++] = x%10;        x/=10;    }    return dfs(pos-1,-1,0,true);}int main(){    int le,ri;    while(~scanf("%d%d",&le,&ri)&&le+ri){        memset(dp,-1,sizeof(dp));        printf("%d\n",solve(ri)-solve(le-1));    }    return 0;}