HDU 3555 Bomb (数位DP)

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

3150500

Sample Output

0115          

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.         

题意：

数出含有49的个数

和here一样

#include <bits/stdc++.h>using namespace std;#define LL long long#define BITNUM 20#define MAXN 10LL dp[BITNUM][MAXN];int bits[BITNUM];///len数字的位数，digit标记状态的值（注意压缩抽象），limit 表示digit是否是第len位（从低位向高位数，个位为第1位）的范围边界LL dfs(int pos, int digit, bool end_flag){    if (!end_flag  && dp[pos][digit] != -1)///记忆化搜索，如果之前已经求出来了，则返回。注意这里要求 end_flag为false        return dp[pos][digit];    if(pos==0) return dp[pos][digit]=1;    int end = end_flag ? bits[pos-1] : 9 ;///如果当前位是边界数字N对应位的最大值，则下一位的范围只能从0到边界数字N的下一位的最大值。否则为0 到 9    LL ans = 0;    for(LL i = 0; i <= end; i++)    {        if(!(digit==4&&i==9))            ans += dfs(pos - 1, i, end_flag && (i==end));    }    if (!end_flag) ///digit不是第len位的最高范围，则可以将结果缓存        dp[pos][digit] = ans;    return ans;}LL solve(LL  x){    memset(bits,0,sizeof bits);    int pos=0;    while(x)    {        bits[pos++]=x%10;        x/=10;    }    return dfs(pos, bits[pos], 1);///为方便当前为设置为0}int main(){    memset(dp,-1,sizeof dp);    int T;    LL n;    scanf("%d",&T);    while(T--)    {       scanf("%lld",&n);       printf("%lld\n",(n+1)-solve(n));    }    return 0;}