A - Bomb (HDU 3555) 数位dp

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本文探讨了一种特殊的时间炸弹计数问题，其中涉及到从1到N的数字序列中寻找包含子序列“49”的数量，以此来计算炸弹的爆炸威力。文章通过一个具体的示例解释了如何计算从1到500的数字中包含“49”子序列的数量，并提供了一个递归算法的实现代码，用于解决更广泛的N值问题。

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Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 25961 Accepted Submission(s): 9845

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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#include<iostream>
#include<cstring>
using namespace std;
#define ll long long
ll dp[30][30], a[100];
ll pos, r, t;
ll dfs(ll pos, ll pre, bool limit) {
    if (pos == -1) return 1;
    if (!limit && dp[pos][pre] != -1) return dp[pos][pre];
    ll up = limit ? a[pos] : 9;
    ll ans = 0;
    for (ll i = 0; i <= up; i++) {
        if (pre == 4 && i == 9) continue;
        ans += dfs(pos - 1, i, limit && i == a[pos]);
    }
    if (!limit) dp[pos][pre] = ans;
    return ans;
}
ll solve(ll x) {
    pos = 0;
    while (x) {
        a[pos++] = x % 10;
        x /= 10;
    }
    return dfs(pos - 1, -1, true);
}
int main() {
    cin >> t;
    memset(dp, -1, sizeof(dp));
    while (t--) {
        cin >> r;
        cout << r - solve(r) + 1 << endl;
    }
    return 0;
}