HDU 3555 Bomb（数位DP）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3555

Bomb

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

Total Submission(s): 16312 Accepted Submission(s): 5961

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

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

【中文题意】问你从0到n这个范围内包含49的数有多少个，具体点的规则看题目。

【思路分析】典型的数位DP题目，具体详解都在AC代码中，一些基础的定义以及某些含义我在不要62这个题目中写过，在此不再赘述，不要62的链接：http://blog.csdn.net/qq_32866009/article/details/53047478。

【AC代码】

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define LL long long

LL dp[20][12],dight[20];

LL dfs(LL pos,LL pre,LL limit)
{
if(pos==-1)
{
return 0;
}
if(!limit&&dp[pos][pre]!=-1)
{
return dp[pos][pre];
}
LL re=0;
LL tmp=limit?dight[pos]:9;
for(LL i=0;i<=tmp;i++)
{
if(pre==4&&i==9&&!limit)//判断是不是到了最后一位
{
re+=(LL)pow(10,pos);
}
else if(pre==4&&i==9&&limit)//到了最后一位的情况
{
for(LL j=pos-1;j>=0;j--)
{
re+=dight[j]*(LL)pow(10,j);
}
re++;
}
else
{
re+=dfs(pos-1,i,limit&&i==tmp);
}
}
if(!limit)
{
dp[pos][pre]=re;
}
return re;
}

LL cal(LL n)
{
LL len=0;
while(n)
{
dight[len++]=n%10;
n/=10;
}
return dfs(len-1,0,1);
}

int main()
{
memset(dp,-1,sizeof(dp));
int t;
scanf("%d",&t);
while(t--)
{
LL n;
scanf("%I64d",&n);
printf("%I64d\n",cal(n));
}
return 0;
}