hdu3555 经典数位dp

/**********************jibancanyang************************** *Author        :jibancanyang *Created Time  : 五  4/15 23:43:25 2016 *File Name     : hdu3555 .cpp *Problem:典型数位dp *Get:典型模型,坑点就是最后枚举数字的时候,前面一旦出现过了49,后面的每一位数字都要和它比它小以为的不含49的数 *组成一个含49的,这算是大体无后效性,但是有一种特殊情况的补救.***********************[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[21][3];int main(void){#ifdef LOCAL    // freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    cin.sync_with_stdio(false);    dp[0][0] = 1;    dp[1][0] = 10, dp[1][1] = 1, dp[1][2] = 0;    for (int i = 2; i <= 20; i++) {        dp[i][0] = 10 * dp[i - 1][0] - dp[i - 1][1];        dp[i][1] = dp[i - 1][0];        dp[i][2] = dp[i - 1][1] + dp[i - 1][2] * 10;    }    cin >> T;    while (T--) {        cin >> n;        n++;        string x = to_string(n);        int m = x.size();        ll ans = 0;        bool flag = false;        for (int j = m; j >= 1; j--) {            int t = x[m - j] - '0';            ans += dp[j - 1][2] * t ;            if (flag) ans += t * dp[j - 1][0];            else if (t > 4) ans += dp[j - 1][1];//注意这里有个else,是为了不要和上一条语句重复加了.            if (t == 9 && j != m && x[m - j - 1] == '4') flag = true;        }        cout << ans << endl;    }    return 0;} *Get: