Bomb

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

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  

Recommend 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;
}