HDU 3555 Bomb

Bomb

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

Total Submission(s): 7907 Accepted Submission(s): 2774



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


解题思路：数位DP，dp[i][0]表示长度为i且不含49的数的个数，dp[i][1]代表长度为i且不含49，但是首位为9的数的个数，dp[i][2]代表长度为i且含49的数的个数
具体状态转移和数字分解见代码

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long dp[22][3];
void DP()
{
    int i;
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(i=1;i<=20;i++)
    {
        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
        dp[i][1]=dp[i-1][0];
        dp[i][2]=dp[i-1][1]+dp[i-1][2]*10;
    }
}
int main()
{
    int i,t,num[20],k,last;
    long long n,ans;
    DP();
    scanf("%d",&t);
    while(t--)
    {
        ans=last=0;k=1;
        memset(num,0,sizeof(num));
        scanf("%I64d",&n);
        n++;
        while(n)
        {
            num[k++]=n%10;
            n/=10;
        }
        k--;
        bool flag=false;
        for(i=k;i>=1;i--)
        {
            ans+=dp[i-1][2]*num[i];
             if(flag)
            {
                ans+=dp[i-1][0]*num[i];
            }
            if(!flag && num[i]>4)
            {
                ans+=dp[i-1][1];
            }
            if(last==4 && num[i]==9)
            {
                flag=true;
            }
            last=num[i];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}