HDU3555 Bomb

　　　　　　　　         Bomb

　　　Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Problem Description

The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

Output

For each test case, output an integer indicating the final points of the power.

Sample Input

3 1 50 500

Sample Output

0 1 15

Hint

From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15.

 

Solution：

　　　　很裸的数位dp，记忆化搜索好啊，不过数组开小了，炸了几次。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
long long T,dp[105][15],bit[105];
#define int long long
long long n;
long long dfs(int pos,int pre,bool Limit) {
    if(pos<=0&&pre>=0) return 1;
    if(pos<=0) return 0;
    if(pre>=0&&!Limit&&dp[pos][pre]!=-1) return dp[pos][pre];
    int End=Limit?bit[pos]:9;
    long long ans=0;
    for(int now,i=0; i<=End; ++i) {
        if(pre==4&&i==9) continue;
        now=i;
        if(pre<0&&!i) now=pre;
        ans+=dfs(pos-1,now,Limit&&(i==End));
    } if(pre>=0&&!Limit) dp[pos][pre]=ans;
    return ans;
}
signed main() {
    //freopen("data.in","r",stdin);
    //freopen("my.out","w",stdout);
    scanf("%lld",&T);
    for(int t=1; t<=T; ++t,bit[0]=0) {
        memset(dp,0xff,sizeof(dp));
        scanf("%lld",&n); 
        long long x=n;
        while(n) bit[++bit[0]]=n%10,n/=10;
        printf("%lld\n",x-dfs(bit[0],-10,true));
    } return 0;
}