hdu3555---Bomb(数位dp,水)

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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这题没用那个经典的板子

设dp[i][0]表示位数小于等于i的所有数中, 不包含49的数的个数
dp[i][1]表示位数小于等于i的所有数中,第i位为9,且不包含49的数的个数
dp[i][2]表示位数小于等于i的所有数中,包含49的数的个数

/*************************************************************************
    > File Name: hdu3555.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月22日 星期日 22时13分31秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

LL dp[70][3];
int bit[70];

void calc (LL n)
{
    int cnt = 0;
    while (n)
    {
        bit[++cnt] = n % 10;
        n /= 10;
    }
    LL ans = 0;
    bool flag = false;
    bit[cnt + 1] = 0;
    for (int i = cnt; i >= 1; --i) //枚举比n开始小的那一位
    {
        ans += dp[i - 1][2] * bit[i]; //先统计i-1位以下的符合条件的数的个数
        if (flag) //既然前缀中出现了49,后面的随意填
        {
            ans += dp[i - 1][0] * bit[i];
        }
        if (!flag && bit[i] > 4) //前缀里没有49,但是此时可以产生49
        {
            ans += dp[i - 1][1];
        }
        if (bit[i + 1] == 4 && bit[i] == 9)
        {
            flag = 1;
        }
    }
    if (flag) //本身
    {
        ++ans;  
    }
    printf("%lld\n", ans);
}

int main ()
{
    dp[0][0] = 1;
    for (int i = 1; i <= 64; ++i)
    {
        dp[i][0] = dp[i - 1][0] * 10 - dp[i - 1][1]; //不包含的情况下要减去第i位恰好为4的情况
        dp[i][1] = dp[i - 1][0];
        dp[i][2] = dp[i - 1][2] * 10 + dp[i - 1][1];
    }
    int t;
    scanf("%d", &t);
    while (t--)
    {
        LL n;
        scanf("%lld", &n);
        calc (n);
    }
    return 0;
}