hdu3555 Bomb(数位dp）

题目传送门

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 23853    Accepted Submission(s): 8990

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

 

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题意：给你n，从[1,n]中找出含有“49”的数的个数

题解：数位dp入门题

代码：

第一份是间接法做的

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
ll n;
int bit[20];
ll dp[20][2];
ll dfs(int pos,int pre,int sta,bool limit)
{
    if(pos==-1)return 1;
    if(!limit&&dp[pos][sta]!=-1)return dp[pos][sta];
    int up=limit?bit[pos]:9;
    ll ans=0;
    for(int i=0;i<=up;i++)
    {
        if(pre==4&&i==9)
            continue;
        ans+=dfs(pos-1,i,i==4,limit&&i==bit[pos]);
    }
    if(!limit)dp[pos][sta]=ans;
    return ans;
}
ll solve(ll x)
{
    int len=0;
    while(x)
    {
        bit[len++]=x%10;
        x/=10;
    }
    return dfs(len-1,-1,0,true);
}
int main()
{
    int T;
     scanf("%d",&T);
     while(T--){
         scanf("%lld",&n);
        memset(dp,-1,sizeof(dp));
        printf("%lld
",n+1-solve(n));
     }
    return 0;
}

 下面的是直接法做的：

  pre=0: 没有49; pre=1: 前一位为4; pre=2: 前几位中有49

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
ll n;
int bit[20];
ll dp[20][3];
ll dfs(int pos,int pre,bool limit)
{
    if(pos==-1)return pre==2;
    if(!limit&&dp[pos][pre]!=-1)return dp[pos][pre];
    int up=limit?bit[pos]:9;
    ll ans=0;
    for(int i=0;i<=up;i++)
    {
        if(pre==2||pre==1&&i==9)
        ans+=dfs(pos-1,2,limit&&i==bit[pos]);
        else if(i==4)
         ans+=dfs(pos-1,1,limit&&i==bit[pos]);
        else
        ans+=dfs(pos-1,0,limit&&i==bit[pos]);
    }
    if(!limit)dp[pos][pre]=ans;
    return ans;
}
ll solve(ll x)
{
    int len=0;
    while(x)
    {
        bit[len++]=x%10;
        x/=10;
    }
    return dfs(len-1,0,true);
}
int main()
{
    int T;
     scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
     while(T--){
         scanf("%lld",&n);
        printf("%lld
",solve(n));
     }
    return 0;
}