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
 
Recommend
zhouzeyong   |   We have carefully selected several similar problems for you:  

pid=3554" style="color:rgb(26,92,200); text-decoration:none">3554 3556 3557 3558 3559 

 
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));
}
}

人家的代码http://blog.csdn.net/scf0920/article/details/42870573

#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;
}

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