-卵の数と最大フロアを入力します。卵の数は 4 で、最大フロアは 10 であるとします。
出力 - 最小試行数 4.
AlgorithmeggTrialCount(egg, Floor)Input- 卵の数、最大フロア。
出力 - 卵テストの最小数を取得します。
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
#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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