1、将一整数逆序后放入一数组中（要求递归实现） void convert(int *result, int n) { if(n>=10) convert(result+1, n/10); *result = n%10; } int main(int argc, char* argv[]) { int n = 123456789, result[20]={}; convert(result, n); printf("%d:", n); for(int i=0; i<9; i++) printf("%d", result[i]); } 2、求高于平均分的学生学号及成绩（学号和成绩人工输入） double find(int total, int n) { int number, score, average; scanf("%d", &number); if(number != 0) { scanf("%d", &score); average = find(total+score, n+1); if(score >= average) printf("%d:%d\n", number, score); return average; } else { printf("Average=%d\n", total/n); return total/n; } } int main(int argc, char* argv[]) { find(0, 0); } 3、递归实现回文判断（如：abcdedbca 就是回文，判断一个面试者对递归理解的简单程序） int find(char *str, int n) { if(n<=1) return 1; else if(str[0]==str[n-1]) return find(str+1, n-2); else return 0; } int main(int argc, char* argv[]) { char *str = "abcdedcba"; printf("%s: %s\n", str, find(str, strlen(str)) ? "Yes" : "No"); } 4、组合问题（从 M 个不同字符中任取 N 个字符的所有组合） void find(char *source, char *result, int n) { if(n==1) { while(*source) printf("%s%c\n", result, *source++); } else { int i, j; for(i=0; source[i] != 0; i++); for(j=0; result[j] != 0; j++); for(; i>=n; i--) { result[j] = *source++; result[j+1] = '\0'; find(source, result, n-1); } } } int main(int argc, char* argv[]) { int const n = 3; char *source = "ABCDE", result[n+1] = {0}; if(n>0 && strlen(source)>0 && n<=strlen(source)) find(source, result, 3); } 5、分解成质因数(如 435234=251*17*17*3*2，据说是华为笔试题) void prim(int m, int n) { if(m>n) { while(m%n != 0) n++; m /= n; prim(m, n); printf("%d*", n); } } int main(int argc, char* argv[]) { int n = 435234; printf("%d=", n); prim(n, 2);...
