這是一個著名的難題。假設有一棟 n 層樓的建築,如果我們有 m 個雞蛋,那麼我們如何找到可以安全地將雞蛋掉落而不打破雞蛋的樓層所需的最少掉落次數。
有一些重要的要點需要記住 -
輸入- 雞蛋數量和最大樓層。假設雞蛋數量為 4,最大樓層為 10。
輸出- 最小試驗次數 4。
#輸入− 雞蛋數量、最大樓層。
輸出 − 取得雞蛋的最小數量試驗。
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
即時示範
#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)); }
Enter number of eggs: 4 Enter max Floor: 10 Minimum number of trials: 4
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