hdu 3555 Bomb （ 數位DP）

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7926 Accepted Submission(s): 2780


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. 

思路： dp[i][0]表示不含49的個數 ；
dp[i][1]表示不含49且高位為9的個數；
dp[i][2]表示包含49的個數；
dp[i][0]=10*dp[i-1][0]-dp[i-1][1]; //不含49的數可以任意加10個數字，減去9前面加4的個數
dp[i][1]=dp[i-1][0]; //不含49的數最高位加9
dp[i][2]=dp[i-1][2]*10+dp[i-1][i]; //包含49的數字可以添加0~9，高位為9的可以加4；

#include"stdio.h"
#include"string.h"
#include"iostream"
#include"algorithm"
#include"math.h"
#include"vector"
using namespace std;
#define LL __int64
#define N 25
LL dp[N][3];
void inti()
{
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    int i;
    for(i=1; i0; i--)
    {
        ans+=a[i]*dp[i-1][2];
        if(flag)
            ans+=a[i]*dp[i-1][0];
        else
        {
            if(a[i]>4)
                ans+=dp[i-1][1];
        }
        if(a[i+1]==4&&a[i]==9)
            flag=1;
        //     printf("%d\n",ans);
    }
    return ans;
}
int main()
{
    int T;
    LL n;
    inti();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d",&n);
        printf("%I64d\n",work(n+1));
    }
    return 0;
}