HDOJ 题目3555 Bomb（数位DP）

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 10419    Accepted Submission(s): 3673

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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这样的比递推的那种写法慢了些啊

ac代码

#include<stdio.h>
#include<string.h>
int bit[20];
__int64 dp[20][10][2];
__int64 dfs(int pos,int pre,int isture,int limit)
{
	if(pos<0)
	{
		return isture;
	}
	if(!limit&&dp[pos][pre][isture]!=-1)
	{
		return dp[pos][pre][isture];
	}
	int last=limit?

bit[pos]:9;
	__int64 ans=0;
	for(int i=0;i<=last;i++)
	{
		ans+=dfs(pos-1,i,isture||(pre==4&&i==9),limit&&(i==last));
	}
	if(!limit)
	{
		dp[pos][pre][isture]=ans;
	}
	return ans;
}
__int64 solve(__int64 n)
{
	int len=0;
	while(n)
	{
		bit[len++]=n%10;
		n/=10;
	}
	return dfs(len-1,0,0,1);
}
int main()
{
	int n;
	memset(dp,-1,sizeof(dp));
	int t;
	scanf("%d",&t);
	while(t--)
	{
		__int64 n;
		scanf("%I64d",&n);
		printf("%I64d\n",solve(n));
	}
}

人家的代码http://blog.csdn.net/scf0920/article/details/42870573

#include <iostream>
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <set>
#include <stdio.h>
using namespace std;
#define LL __int64
#define pi acos(-1.0)
const int mod=100000000;
const int INF=0x3f3f3f3f;
const double eqs=1e-8;
LL dp[21][11], c[21];
LL dfs(int cnt, int pre, int maxd, int zero)
{
    if(cnt==-1) return 1;
    if(maxd&&zero&&dp[cnt][pre]!=-1) return dp[cnt][pre];
    int i, r;
    LL ans=0;
    r=maxd==0?

c[cnt]:9;
    for(i=0;i<=r;i++){
        if(!zero||!(pre==4&&i==9)){
            ans+=dfs(cnt-1,i,maxd||i<r,zero||i);
        }
    }
    if(maxd&&zero) dp[cnt][pre]=ans;
    return ans;
}
LL Cal(LL x)
{
    int i, cnt=0;
    while(x){
        c[cnt++]=x%10;
        x/=10;
    }
    return dfs(cnt-1,-1,0,0);
}
int main()
{
    int t;
    LL n;
    scanf("%d",&t);
    memset(dp,-1,sizeof(dp));
    while(t--){
        scanf("%I64d",&n);
        printf("%I64d\n",n+1-Cal(n));
    }
    return 0;
}