hdu3555 Bomb(数位dp)

                                                   Bomb

                                                                                    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
                                                                                                             Total Submission(s): 18060    Accepted Submission(s): 6639

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

推荐一份题解 写的很好  点我

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<cstring>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<algorithm>
#include<string.h>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=0.0000000001;
const int N=100;
ll dp[N][N];
int a[N];
void init(){
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;dp[0][1]=0;dp[0][2]=0;
    for(int 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 T;
    init();
    scanf("%d",&T);
    while(T--){
        memset(a,0,sizeof(a));
        ll n;
        scanf("%I64d",&n);
        n++;
        int t=0;
        while(n){
            a[++t]=n%10;
            n/=10;
        }
        a[t+1]=0;
        int flag=0;
        ll sum=0;
        for(int i=t;i>=1;i--){
            sum=sum+dp[i-1][2]*a[i];
            if(flag==0&&a[i]>4)  sum=sum+dp[i-1][1];
            if(flag)    sum=sum+dp[i-1][0]*a[i];
            if(a[i+1]==4&&a[i]==9)  flag=1;
        }
        cout<<sum<<endl;
    }
    return 0;
}