HDU 3555 Bomb（数位DP模板啊两种形式）

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

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

题意：

求0 到n的数中有多少个数字是含有‘49’的。

PS：

数位DP

//dp[i][j]:长度为i的数的第j种状态

//dp[i][0]:长度为i可是不包括49的方案数

//dp[i][1]:长度为i且不含49可是以9开头的数字的方案数

//dp[i][2]:长度为i且包括49的方案数

(转)状态转移例如以下

dp[i][0] = dp[i-1][0] * 10 - dp[i-1][1];  // not include 49  假设不含49且，在前面能够填上0-9 可是要减去dp[i-1][1] 由于4会和9构成49

dp[i][1] = dp[i-1][0];  // not include 49 but starts with 9  这个直接在不含49的数上填个9即可了

dp[i][2] = dp[i-1][2] * 10 + dp[i-1][1]; // include 49  已经含有49的数能够填0-9，或者9开头的填4



接着就是从高位開始统计



在统计到某一位的时候，加上 dp[i-1][2] * digit[i] 是显然对的。由于这一位能够填 0 - (digit[i]-1)

若这一位之前挨着49，那么加上 dp[i-1][0] * digit[i] 也是显然对的。

若这一位之前没有挨着49，可是digit[i]比4大，那么当这一位填4的时候，就得加上dp[i-1][1]

代码例如以下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
LL dp[27][3];
int c[27];
//dp[i][j]:长度为i的数的第j种状态
//dp[i][0]:长度为i可是不包括49的方案数
//dp[i][1]:长度为i且不含49可是以9开头的数字的方案数
//dp[i][2]:长度为i且包括49的方案数
void init()
{
    memset(dp,0,sizeof(dp));
    dp[0][0] = 1;
    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]*1;
        dp[i][2] = dp[i-1][2]*10+dp[i-1][1];
    }
}

int cal(LL n)
{
    int k = 0;
    memset(c,0,sizeof(c));
    while(n)
    {
        c[++k] = n%10;
        n/=10;
    }
    c[k+1] = 0;
    return k;
}
void solve(int len, LL n)
{
    int flag = 0;//标记是否出现过49
    LL ans = 0;
    for(int i = len; i >= 1; i--)
    {
        ans+=c[i]*dp[i-1][2];
        if(flag)
        {
            ans+=c[i]*dp[i-1][0];
        }
        else if(c[i] > 4)
        {
            //这一位前面没有挨着49。但c[i]比4大，那么当这一位填4的时候，要加上dp[i-1][1]
            ans+=dp[i-1][1];
        }
        if(c[i+1]==4 && c[i]==9)
        {
            flag = 1;
        }
    }
    printf("%I64d\n",ans);
}
int main()
{
    int t;
    LL n;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d",&n);
        int len = cal(n+1);
        solve(len, n);
    }
    return 0;
}

DFS版

代码例如以下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL __int64
LL n, dp[25][3];
//dp[i][j]:长度为i。状态为j
int digit[25];
//nstatus: 0：不含49, 1：不含49但末尾是4, 2 :含49
LL DFS(int pos, int status, int limit)
{
    if(pos <= 0) // 假设到了已经枚举了最后一位。而且在枚举的过程中有49序列出现
        return status==2;//注意是 ==
    if(!limit && dp[pos][status]!=-1)   //对于有限制的询问我们是不可以记忆化的
        return dp[pos][status];
    LL ans = 0;
    int End = limit?digit[pos]:9;   // 确定这一位的上限是多少
    for(int i = 0; i <= End; i++)   // 每一位有这么多的选择
    {
        int nstatus = status;       // 有点else s = statu 的意思

        if(status==0 && i==4)//高位不含49。而且末尾不是4 ，如今末尾添4返回1状态
            nstatus = 1;
        else if(status==1 && i!=4 && i!=9)//高位不含49。且末尾是4，如今末尾加入的不是4返回0状态
            nstatus = 0;
        else if(status==1 && i==9)//高位不含49，且末尾是4，如今末尾加入9返回2状态
            nstatus = 2;
        ans+=DFS(pos-1, nstatus, limit && i==End);
    }
    if(!limit)
        dp[pos][status]=ans;
    return ans;
}

int cal(LL x)
{
    int cnt = 0;
    while(x)
    {
        digit[++cnt] = x%10;
        x/=10;
    }
    digit[cnt+1] = 0;
    return cnt;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,-1,sizeof(dp));
        scanf("%I64d",&n);
        int len = cal(n);
        LL ans = DFS(len, 0, 1);
        printf("%I64d\n",ans);
    }
    return 0;
}