数据结构基础(C语言版)FUNDAMENTALSOFDATASTRUCTURESINC部分习题英文版答案..docx

本人答案整理自/~sahni/fdsc2ed/没有进行校对,仅为方便自己断网或网络不好时使用.如有纰漏,望请见谅.因见多有人寻此资源不果,遂分享之,以报多年来众驴友分享于我之万一.本答案为英文版数据结构基础(C语言版)即FUNDAMENTALS OF DATA STRUCTURES IN C 部分习题答案.答案所有文字描述均为英文.且仅有大部分习题答案.少数习题答案网站没有给出,因而我也没有办法.本人排版水平不佳,全部复制粘贴网站内容,几乎未经过任何排版,大家就将就看下吧.CHAPTER1Chapter 1, Pages 16-17Problem 1.a This statement is a poorly phrased version of Fermats last theorem. We know that we can find n 2 for which ?the equation holds. Fermat scribbled a note on a text margin indicating that he had found the solution for n = 2. Unfortunately, the proof died with him. Because the statement is phrased as a question? rather than a clearly defined algorithm, it lacks definiteness.Problem 1.b This statement violates not only the rules of mathematics, but the criterion of effectiveness. We can compute only those things that are feasible, and division by zero is mathematically undefinedPage 17, Exercise 3#include stdio.h#include math.h#include string.h#define TRUE 1#define FALSE 0#define MAX_STRING 100void truth_table(int);int main(){ int n; printf(n:(=0): ); scanf(%d, n); while (n = 0) { /*error loop */ printf(n:(=0): ); scanf(%d, n); } truth_table(n);}void truth_table(int n){/* generate a truth_table by transforming # of permutations into binary */int i,j, div, rem;char string[MAX_STRING];for (i=0; i pow(2,n); i++){/*number of permutations or rows in the table */ strcpy(string ,\0); div = i; for (j = n; j 0; j--) {/*number of bits needed for each row*/ rem = div%2; div = div/2; if (!rem) strcat(string,FALSE ); else strcat(string, TRUE ); } printf(%s\n, string); }}Page 17, Exercise 4#include stdio.hint min(int, int);#define TRUE 1#define FALSE 0int main(){ int x,y,z; printf(x: ); scanf(%d, x); printf(y: ); scanf (%d, y); printf(z: ); scanf(%d, z); if (min(x,y) min(y,z)) {/*x is smallest */ printf(%d , x); if (min(y,z)) printf (%d %d\n, y,z); else printf(%d%d\n, z, y);} else if (min(y,x) min(y,z)){ /*y is the smallest */ printf(%d , y);