HDU 3555 Bomb（数位DP啊）

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

[email protected]

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;
}

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