这是一个著名的难题。假设有一栋 n 层楼的建筑，如果我们有 m 个鸡蛋，那么我们如何找到可以安全地将鸡蛋掉落而不打破鸡蛋的楼层所需的最少掉落次数。
有一些重要的要点需要记住 -
- 当鸡蛋没有从给定的楼层破裂时，那么它在任何较低的楼层也不会破裂。
- 如果一个鸡蛋从给定的楼层破裂，那么它也会在所有上面的楼层破裂。
- 当鸡蛋破裂时，它必须被丢弃，否则我们可以再次使用它。
输入- 鸡蛋数量和最大楼层。假设鸡蛋数量为 4，最大楼层为 10。
输出- 最小试验次数 4。
算法
eggTrialCount(鸡蛋，楼层)
输入− 鸡蛋数量、最大楼层。
输出 − 获取鸡蛋的最小数量试验。
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
实时演示
#include<stdio.h>
#define MAX_VAL 9999
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 i-th egg
and jth floor
int res, i, j, k;
for (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 (j = 1; j <= floors; j++) //when egg is 1, we need 1 trials for
each floor
minTrial[1][j] = j;
for (i = 2; i <= eggs; i++){ //for 2 or more than 2 eggs
for (j = 2; j <= floors; j++) { //for second or more than second
floor
minTrial[i][j] = MAX_VAL;
for (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;
printf("Enter number of eggs: ");
scanf("%d", &egg);
printf("Enter max Floor: ");
scanf("%d", &maxFloor);
printf("Minimum number of trials: %d", eggTrialCount(egg, maxFloor));
}输出
Enter number of eggs: 4 Enter max Floor: 10 Minimum number of trials: 4
