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 -
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。
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
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)); }
Enter number of eggs: 4 Enter max Floor: 10 Minimum number of trials: 4
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