HDU 2089

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http://acm.hdu.edu.cn/showproblem.php?pid=2089

这是个简单的数位DP，具体看代码吧、

#include <cstdio>#include <cstring>#include <string>#include <queue>#include <stack>#include <vector>#include <iostream>#include <algorithm>using namespace std;#define mod 1000000007#define ll long long#define M 20int num[M] , dp[M];//num存放数字的每一位，dp存放在没有限制的条件下当前数字长度满足条件的个数int Dfs(int index , int fg , int lim)//fg表示上一位是否为6，lim表示上限是否受限。{if (!index)return 1;if (!lim && !fg && dp[index] != -1)return dp[index];int i , ret = 0 , up = lim ? num[index] : 9;for (i = 0 ; i <= up ; i++){if (i == 4)continue;if (fg && i == 2)continue;ret += Dfs(index-1 , i==6 , lim && i == up);}if (!lim && !fg)dp[index] = ret;return ret;}int Solve(int k)//分离出每一位数字{int len = 0;while (k){num[++len] = k%10;k /= 10;}return Dfs(len,0,1);}int main(){int m , n;memset(dp , -1 , sizeof dp);while (scanf("%d%d",&m,&n),m){printf("%d\n",Solve(n)-Solve(m-1));}return 0;}