hdu3555 模板化数位dp

/**********************jibancanyang************************** *Author        :jibancanyang *Created Time  : 五  4/15 23:43:25 2016 *File Name     : hdu3555 .cpp *Problem:典型数位dp *Get:上次切这个题用的是水过去的方法,这次用模板化的dfs写了下,其方法本质就是一位一位的增加数字. *并且设置了一个limit的状态记录,来记录前面的数是否是边界,如果是边界那么当前位只能最大是bit[i],而不能 *0 ~ 9的随便取了.然后注意的是这里是统计的到0的情况.***********************[e-mail]**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define rep(i, a, n) for (int i = a; i < n; i++)#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3fffffff, maxn = 1e5 + 12;int T;ll n, dp[25][2];int bits[25];#define limit asdfll dfs(int len, bool four, bool limit) {    if (len == 0) return 1;    if (!limit && dp[len][four] != -1) return dp[len][four];    int m = limit ? bits[len] : 9;    ll ret = 0;    for (int i = 0; i <= m; i++) {        if (four && i == 9) continue; //避免生成49        ret += dfs(len - 1, i == 4, limit && i == m);    }    //pr(len), pr(four), pr(limit), pl(ret);     if (!limit) dp[len][four] = ret;    return ret;}ll solve(ll n) {    ll key = n, t = 1;    while (key) {        bits[t++] = key % 10;        key /= 10;    }    return dfs(t - 1, false, true);}int main(void){#ifdef LOCAL    // freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    cin.sync_with_stdio(false);    cin >> T;    memset(dp, -1, sizeof(dp));    while (T--) {        cin >> n;        cout << n - solve(n)  + 1 << endl;//+1是因为00000    }    return 0;}