HDU3555——Bomb（数位dp入门）

Bomb

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

Total Submission(s): 15805    Accepted Submission(s): 5761


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

HintFrom 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

 

Recommend

zhouzeyong

之前零零碎碎写过几道数位dp的题目，但遇到了自己还是写不出来。决定系统做一做啦。从入门题开始。

第一道题跟着kuangbin巨巨的模板写的。加了一些自己的注释。

#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <ctime>
#include <cstdlib>
#include <cctype>
#include <iostream>
#include <fstream>
using namespace std;
#define MAXN 400010
#define LEN 200010
#define INF 1e9+7
#define MODE 1000000
#define pi acos(-1)
#define g 9.8
typedef long long ll;

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用来储存每一位的值分别为多少
bit[++len]=n%10;
n/=10;
}
bit[len+1]=0;
bool flag=false;
long long ans=0;
for(int i=len;i>=1;i--)
{
//从最高位开始处理
//首先是如果后面的位数包括49，那么最高位随便取。
ans+=dp[i-1][2]*bit[i];
//也就说存在更高位为包括49的情况
if(flag)ans+=dp[i-1][0]*bit[i];
else
{
//如果下一位为9，且当前位大于4，则存在包括49的情况
if(bit[i]>4)ans+=dp[i-1][1];
}
//flag=true最大值的上一位和当前位是49，之后flag的值不会再改变。
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;
}