Bomb HDU - 3555 (数位dp)

最新推荐文章于 2025-08-07 15:32:36 发布

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最新推荐文章于 2025-08-07 15:32:36 发布

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分类专栏：数位dp 文章标签： dp 算法 dfs

本文链接：https://blog.youkuaiyun.com/zzcblogs/article/details/70504413

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本文介绍了一个算法问题，即计算从1到N内包含子序列'49'的数字数量，以此来确定时间炸弹的最终威力点数。文章通过递归深度优先搜索结合记忆化的实现方式，高效地解决了这一问题。

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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

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.

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <map>
#include <vector>
using namespace std;
int a[25];
long long dp[25][2][12];//当前位数，是否已包含49，前一位是几
    //      当前位置     是否包含49。  前一位      限制位
long long dfs(int pos, bool have, int pre,  bool limit) {
        //如果已枚举完，并且已经包含49 则输出1，否则0了。
    if (pos == -1) return have;
        //记忆化dp的正确姿势
    if (!limit && dp[pos][have][pre] != -1) return dp[pos][have][pre];
    int up = limit ? a[pos] : 9;
    long long temp = 0;
    
    for (int i = 0; i <= up; ++i) {
        if (pre == 4 && i == 9) {//存在49的状态
            temp += dfs(pos-1, true, i, limit && i == a[pos]);
        } else {                 //不存在49的状态
            temp += dfs(pos-1, have, i, limit && i == a[pos]);
        }
    }
    
    if (!limit)
        dp[pos][have][pre] = temp;
    return temp;
}
long long solve(long long x) {
    int i = 0;
    while (x) {
        a[i++] = x%10;
        x /= 10;
    }
    return dfs(i-1, false, 0, 1);
}
int main() {
    int T;
    long long shit;
    cin >> T;
    while (T--) {
        memset(dp, -1, sizeof(dp));
        cin >>shit;
        cout << solve(shit) << endl;
    }
}