hdu 3555 Bomb（数位dp）

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此博客讨论了从1到N的范围内找到包含特定子序列的数的个数问题，涉及数学逻辑与编程实现。

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Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 4054 Accepted Submission(s): 1401

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

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

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zhouzeyong

题意：找1~N的包含49的数的个数

#include<stdio.h>
#include<string.h>
long long dp[28][3],n,res;
int c[28];
int main()
{
    int i,t,flag,len;

    memset(dp,0,sizeof(dp));
    for(dp[0][0]=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;
    }
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d",&n);
        memset(c,0,sizeof(c));
        res=flag=len=0;
        n++;
        while(n)
        {
            c[++len]=n%10;
            n/=10;
        }
        for(i=len;i>=1;i--)
        {
            res+=dp[i-1][2]*c[i];
            if(flag) res+=dp[i-1][0]*c[i];
            if(!flag&&c[i]>4) res+=dp[i-1][1];
            if(c[i+1]==4&&c[i]==9) flag=1;
        }
        printf("%I64d\n",res);
    }

    return 0;
}