HDU 3555 Bomb（数位DP）

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 3362 Accepted Submission(s): 1185


[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, output an integer indicating the final points of the power.

[align=left]Sample Input[/align]

3
1
50
500

[align=left]Sample Output[/align]

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.

[align=left]Author[/align]
fatboy_cw@WHU

[align=left]Source[/align]
2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

[align=left]Recommend[/align]
zhouzeyong

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

简单的数位DP入门题。
注释见代码：

/*
* HDU 3555
* 求1~N中含有数字49的个数     1 <= N <= 2^63-1
*/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
long long dp[25][3];
/*
* dp[i][0],表示不含有49
* dp[i][1],表示不含有49，且最高位为9
* dp[i][2],表示含有49
*/
void init()
{
dp[0][0]=1;
dp[0][1]=dp[0][2]=0;
for(int i=1;i<25;i++)
{
dp[i][0]=10*dp[i-1][0]-dp[i-1][1];//在前面加0~9的数字，减掉在9前面加4
dp[i][1]=dp[i-1][0];//最高位加9
dp[i][2]=10*dp[i-1][2]+dp[i-1][1];//在本来含有49的前面加任意数，或者在9前面加4
}
}
int bit[25];
long long calc(long long n)
{
int len=0;
while(n)
{
bit[++len]=n%10;
n/=10;
}
bit[len+1]=0;
bool flag=false;
long long ans=0;
for(int i=len;i>=1;i--)
{
ans+=dp[i-1][2]*bit[i];
if(flag)ans+=dp[i-1][0]*bit[i];
else
{
if(bit[i]>4)ans+=dp[i-1][1];
}
if(bit[i+1]==4&&bit[i]==9)flag=true;
}
if(flag)ans++;//加上n本身
return ans;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
long long n;
scanf("%d",&T);
init();
while(T--)
{
scanf("%I64d",&n);
printf("%I64d\n",calc(n));
}
return 0;
}