hdu 3555 Bomb 【数位DP】

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3555

题意：上一题是不要62 这个是“不要49”

代码：

#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <ctime>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <assert.h>using namespace std;#define N 50using namespace std;int bit[N];long long dp[N][3];/*      dp[i][0]:前i位不含 49 的个数。　　  dp[i][1]:前i位不含 49 数且i+1位是4的个数。  　　dp[i][2]:前i位含 49 的个数。*/long long dfs(int pos, int st, bool flag) {    if (pos == 0) return st == 2;    if (flag && dp[pos][st] != -1) return dp[pos][st];    long long ans = 0;    int u = flag ? 9 : bit[pos];     for (int d = 0;d <= u;d++)     {        if (st == 2 || (st == 1 && d == 9))            ans += dfs(pos - 1, 2, flag || d<u);        else if (d == 4)            ans += dfs(pos - 1, 1, flag || d<u);        else             ans += dfs(pos - 1, 0, flag || d<u);    }    if (flag) dp[pos][st] = ans;    return ans;}long long solve(long long n) {    int len = 0;    while (n)     {        bit[++len] = n % 10;        n /= 10;    }    return dfs(len, 0, 0);}int main() {    long long n;    int t;    scanf("%d",&t);    while (t--)     {        scanf("%lld", &n);        memset(dp, -1, sizeof(dp));        printf("%lld\n", solve(n));    }    return 0;}

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