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法律研究扔雞蛋問題
這是一道著名的難題。假設有一棟 n 層樓的建築，如果我們有 m 個雞蛋，那麼如何找出最少需要幾次才能找到一個可以安全扔下雞蛋而不破的地板。
這裡有一些需要記住的重要點 −
- 如果雞蛋從某一層樓扔下時沒有破，那麼它也不會在任何較低層樓上破。
- 如果雞蛋從某一層樓扔下時破了，那麼它會在所有較高的層樓上破。
- 如果雞蛋破了，它必須被丟棄，否則我們可以重複使用它。
輸入和輸出
Input: The number of eggs and the maximum floor. Say the number of eggs are 4 and the maximum floor is 10. Output: Enter number of eggs: 4 Enter max Floor: 10 Minimum number of trials: 4
演算法
eggTrialCount(eggs, floors)
輸入：雞蛋數量，最高樓層。
輸出 − 找出最少次數。
Begin define matrix of size [eggs+1, floors+1] for i:= 1 to eggs, do minTrial[i, 1] := 1 minTrial[i, 0] := 0 done for j := 1 to floors, do minTrial[1, j] := j done for i := 2 to eggs, do for j := 2 to floors, do minTrial[i, j] := ∞ for k := 1 to j, do res := 1 + max of minTrial[i-1, k-1] and minTrial[i, j-k] if res < minTrial[i, j], then minTrial[i,j] := res done done done return minTrial[eggs, floors] End
示例
#include<iostream>
using namespace std;
int max(int a, int b) {
return (a > b)? a: b;
}
int eggTrialCount(int eggs, int floors) { //minimum trials for worst case
int minTrial[eggs+1][floors+1]; //to store minimum trials for ith egg and jth floor
int res;
for (int i = 1; i <= eggs; i++) { //one trial to check from first floor, and no trial for 0th floor
minTrial[i][1] = 1;
minTrial[i][0] = 0;
}
for (int j = 1; j <= floors; j++) //when egg is 1, we need 1 trials for each floor
minTrial[1][j] = j;
for (int i = 2; i <= eggs; i++) { //for 2 or more than 2 eggs
for (int j = 2; j <= floors; j++) { //for second or more than second floor
minTrial[i][j] = INT_MAX;
for (int k = 1; k <= j; k++) {
res = 1 + max(minTrial[i-1][k-1], minTrial[i][j-k]);
if (res < minTrial[i][j])
minTrial[i][j] = res;
}
}
}
return minTrial[eggs][floors]; //number of trials for asked egg and floor
}
int main () {
int egg, maxFloor;
cout << "Enter number of eggs: "; cin >> egg;
cout << "Enter max Floor: "; cin >> maxFloor;
cout << "Minimum number of trials: " << eggTrialCount(egg, maxFloor);
}輸出
Enter number of eggs: 4 Enter max Floor: 10 Minimum number of trials: 4
更新於： 16-Jun-2020
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