Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 10419    Accepted Submission(s): 3673


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.
 

Author
fatboy_cw@WHU
 

Source
 

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这样的比递推的那种写法慢了些啊
ac代码
#include<stdio.h>
#include<string.h>
int bit[20];
__int64 dp[20][10][2];
__int64 dfs(int pos,int pre,int isture,int limit)
{
	if(pos<0)
	{
		return isture;
	}
	if(!limit&&dp[pos][pre][isture]!=-1)
	{
		return dp[pos][pre][isture];
	}
	int last=limit?

 

bit[pos]:9; __int64 ans=0; for(int i=0;i<=last;i++) { ans+=dfs(pos-1,i,isture||(pre==4&&i==9),limit&&(i==last)); } if(!limit) { dp[pos][pre][isture]=ans; } return ans; } __int64 solve(__int64 n) { int len=0; while(n) { bit[len++]=n%10; n/=10; } return dfs(len-1,0,0,1); } int main() { int n; memset(dp,-1,sizeof(dp)); int t; scanf("%d",&t); while(t--) { __int64 n; scanf("%I64d",&n); printf("%I64d\n",solve(n)); } }


#include <iostream>  
#include <string.h>  
#include <math.h>  
#include <queue>  
#include <algorithm>  
#include <stdlib.h>  
#include <map>  
#include <set>  
#include <stdio.h>  
using namespace std;  
#define LL __int64  
#define pi acos(-1.0)  
const int mod=100000000;  
const int INF=0x3f3f3f3f;  
const double eqs=1e-8;  
LL dp[21][11], c[21];  
LL dfs(int cnt, int pre, int maxd, int zero)  
{  
    if(cnt==-1) return 1;  
    if(maxd&&zero&&dp[cnt][pre]!=-1) return dp[cnt][pre];  
    int i, r;  
    LL ans=0;  
    r=maxd==0?

 

c[cnt]:9; for(i=0;i<=r;i++){ if(!zero||!(pre==4&&i==9)){ ans+=dfs(cnt-1,i,maxd||i<r,zero||i); } } if(maxd&&zero) dp[cnt][pre]=ans; return ans; } LL Cal(LL x) { int i, cnt=0; while(x){ c[cnt++]=x%10; x/=10; } return dfs(cnt-1,-1,0,0); } int main() { int t; LL n; scanf("%d",&t); memset(dp,-1,sizeof(dp)); while(t--){ scanf("%I64d",&n); printf("%I64d\n",n+1-Cal(n)); } return 0; }