HDOJ 题目3555 Bomb（数位DP）

最新推荐文章于 2019-05-03 09:27:45 发布

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最新推荐文章于 2019-05-03 09:27:45 发布

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此题为算法挑战，需计算从1到N内包含特定子序列“49”的次数，以此确定炸弹的最终威力值。

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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

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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zhouzeyong | We have carefully selected several similar problems for you: 3554 3556 3557 3558 3559

这种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.youkuaiyun.com/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;  
}