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

3150500

 

Sample Output

0115
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但是不包含49的方案数
//dp[i][1]:长度为i且不含49但是以9开头的数字的方案数
//dp[i][2]:长度为i且包含49的方案数

(转)状态转移如下
dp[i][0] = dp[i-1][0] * 10 - dp[i-1][1];  // not include 49  如果不含49且，在前面可以填上0-9 但是要减去dp[i-1][1] 因为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就行了
dp[i][2] = dp[i-1][2] * 10 + dp[i-1][1]; // include 49  已经含有49的数可以填0-9，或者9开头的填4

接着就是从高位开始统计

在统计到某一位的时候，加上 dp[i-1][2] * digit[i] 是显然对的，因为这一位可以填 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，但是digit[i]比4大，那么当这一位填4的时候，就得加上dp[i-1][1]

代码如下：

#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但是不包含49的方案数//dp[i][1]:长度为i且不含49但是以9开头的数字的方案数//dp[i][2]:长度为i且包含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

代码如下：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define LL __int64LL n, dp[25][3];//dp[i][j]:长度为i，状态为jint digit[25];//nstatus: 0：不含49, 1：不含49但末尾是4, 2 :含49LL DFS(int pos, int status, int limit){    if(pos <= 0) // 如果到了已经枚举了最后一位，并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能够记忆化的        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，并且末尾不是4 ，现在末尾添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，现在末尾添加的不是4返回0状态            nstatus = 0;        else if(status==1 && i==9)//高位不含49，且末尾是4，现在末尾添加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;}