HDU3555 Bomb (digital DP) __ algorithm

The Digital DP Entry topic (I'll just do the entry-level topic 0 0) and find out how many numbers within n don't contain a contiguous 49 number. Similar to another question http://acm.hdu.edu.cn/showproblem.php?pid=2089, copy the code. Another problem, n long ago: transmission door.

Pre-preprocessing, set DP[I][J] represents the number of schemes (J can be 0) for the number of J, and it is not difficult to get dp[i][j] = Sigma (dp[i-1][k)) (0<=k<=9,k≠4 and J≠9).

Simulate the summation of the principle, assuming n = 23495, then len = 5; The first round is to find out the total scheme number of 1xxxx and 0xxxx, that is, the total number of schemes less than 20000; the second cycle is to find the total number of 20xxx and 21xxx and 22xxx; The third time to find out 230xx,231xx,232xx,233xx; 2340x,2341x,2342x,2343x,2344x,2345x,2346x,2347x,2348x, the next is 2349x, This time there are 49 and then there is no meaning, so the cycle of termination.

#include <iostream> #include <cstring> using namespace std;
typedef long Long LL;
LL DP[25][10];
int digit[20];
    void Init () {dp[0][0] = 1;
            for (int i = 1; I <= i++) {for (int j = 0; J <= 9; j +) for (int k = 0; k <= 9; k++)
                {if (k = = 9&&j = 4) continue;
            DP[I][J] + = dp[i-1][k];
    }} void Dig (LL x,int &len) {memset (digit,0,sizeof digit);
        while (x) {Digit[++len] = x%10;
    x/= 10;
    } ll Cal (ll N) {ll ans = 0;
    int len = 0;
    Dig (N,len); for (int i = len; I >= 1; i--) {for (int j = 0; J < Digit[i]; J +) {ans = d
        P[I][J];
    } if (digit[i] = = 9&&digit[i+1] = = 4) break;
return ans;
    int main () {init ();
    int T;
    LL N;
    cin>>t;
        while (t--) {cin>>n;
    Cout<<n+1-cal (n+1) <<endl; }
}