HDU 3555 Bomb（數位DP模板）

http://acm.hdu.edu.cn/showproblem.php?pid=3555

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 25552    Accepted Submission(s): 9677

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 Inpu

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.

基礎數位dp，

題意：給出一個數n，求【1，n】中含有49的數有多少個

數據最大可爲2^63-1

 dp打表，打一個10^19的表，，，

然後照表查詢 ，注意一下查詢方法

 

 

 

打表：

int init()
{
    a[0][0]=1;
    for(int i = 1 ;i < 22 ; i++){///遍歷長度爲i的數字

        for(int j = 0 ; j < 10 ; j++){///長度爲i且開頭爲j的數字

            for(int k = 0 ; k < 10 ; k++){///加上i-1中的個數

                if(j==4&&k==9) continue;
                    a[i][j] += a[i-1][k];
            }
        }
    } 
}

 

AC代碼

#include<bits/stdc++.h>
using namespace std;
#define ll  unsigned long long 
ll a[20][20]; 
int init()
{
    a[0][0]=1;
    for(int i = 1 ;i < 22 ; i++){///遍歷長度爲i的數字

        for(int j = 0 ; j < 10 ; j++){///長度爲i且開頭爲j的數字

            for(int k = 0 ; k < 10 ; k++){///加上i-1中的個數

                if(j==4&&k==9) continue;
                    a[i][j] += a[i-1][k];
            }
        }
    } 
}
int b[20];
int init2(ll n)
{
    int len=1;
    while(n)
    {
        b[len++]=n%10;
        n/=10;
    }
    return len-1;
}
ll solve(ll x)
{
    int len=init2(x);
    ll ans=0;
    for(int i=len;i>0;i--)
    { 
        for(int j=0;j<b[i];j++)
        { 
                ans=ans+a[i][j]; 

        } 
        if(b[i]==9&&b[i+1]==4) break;
    }return ans;
}

int main()
{
   ll t;
    init();
    t=0;
    scanf("%llu",&t); 
    while(t--){
        ll x; 
         scanf("%llu",&x);
        x++;
        memset(b,0,sizeof(b));
        printf("%llu\n",x-solve(x));


    }
    return 0;
}
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