ZSTUOJ 4214: Power Eggs(经典动态规划——鹰蛋问题)

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本文探讨了一个经典的算法问题——鹰蛋问题，并提供了一种基于动态规划的解决方案。问题设定在一个有N层的建筑中，利用K个相同的鹰蛋找出最低的楼层F，使得鹰蛋从该楼层及以下不会破碎，而从该楼层以上则会破碎。文章详细解释了如何通过DP方法找到确定F所需的最少投掷次数。

4214: Power Eggs

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 125 Solved: 21

Description

Benedict bought K identical power eggs from Dropeggs.com, and now he wants to test them by dropping them from different floors of his building. His building has N floors numbered 1 to N. F is an unknown number in the range from 0 to N, inclusive. Each egg will break if dropped from floor F+1 or above, but will not break if dropped from floor F or below. Benedict can drop each egg as many times as he wants from any floor until it breaks. He wants to know the minimum number of egg drops necessary to ensure that he can determine F.

For example, if there are three floors and Benedict has only one egg, then he has to first throw the egg from the first floor, then from the second floor (if the egg survived), and then from the third floor (if the egg survived). Therefore, three drops are required in the worst case

Input

The first line contains one number T (1 ≤ T ≤ 10000) which is the number of test cases, followed by T lines. Each of the next T lines contains two numbers: N, the number of floors (1 ≤ N ≤ 2000000007) and K, the number of eggs (1 ≤ K ≤ 32).

Output

For each of the T lines, print the minimal number of drops required, or if it's greater than 32, print the word Impossible. After that many drops, Benedict gets too tired and cannot continue.

Sample Input

4
10 1
100 2
30 30
2000000000 2

Sample Output

10
14
5
Impossible

鹰蛋问题点击打开链接

经典的鹰蛋问题，首先感谢金巨带我入坑，废掉大量脑细胞之后好不容易看懂点了，

题意：n层楼,k个鸡蛋，求可以测出鸡蛋的最小脆弱度（其实就是鸡蛋最坏条件下可以在哪层楼摔碎）

方法：由于最多可以比较32次，所以只需要开一个dp[32][32]的数组，接着就是恶心的dp思路了。。。。。

设dp[i][j]为i个蛋，j层楼时的鸡蛋的脆弱度

由鹰蛋问题可以知道，dp[1][j] = i, dp[i][1] = 1;

Dp[i][j] = Dp[i][j-1] + Dp[i-1][j-1] + 1;

#include <bits/stdc++.h>
#define LL long long
using namespace std;

LL dp[33][33];//i层楼j个蛋确定的最小次数

void Init()
{
    for(int i=1;i<=32;i++)
        dp[1][i] = i, dp[i][1] = 1;
    for(int i=2;i<=32;i++)
        for(int j=2;j<=32;j++)
            dp[i][j] = dp[i][j-1] + dp[i-1][j-1] + 1;
}
int main()
{
    int T,K;
    LL N;
    Init();
    scanf("%d",&T);
    while(T-- && scanf("%lld %d",&N,&K))
    {
        int pos = -1;
        for(int i=1;i<=32;i++)
        {
            if(dp[K][i] >= N)
            {
                pos = i;
                break;
            }
        }
        if(pos!=-1) printf("%d\n",pos);
        else printf("Impossible\n");
    }
    return 0;
}