2006年中科大计算机考研复试机试题

这是2006年USTC的机试试题，试题来源于网络，解答为我的原创。转载请注明出处。谢谢~

01.读int矩阵文件,将之转置后输出

// 简单，无需解释#include <stdio.h> int main() { int m, n; //A[m][n] int A[100][100]; int i, j; while (scanf ("%d %d", &m, &n) != EOF) { for (i = 0; i < m; i++) //读入矩阵 for (j = 0; j < n; j++) scanf ("%d", &A[i][j]); for (i = 0; i < n; i++) //输出逆转的矩阵 { for (j = 0; j < m; j++) printf ("%d ", A[j][i]); printf ("\n"); } } return 0; }

02.给出时间格式 [年,月], 判断此月有多少天.

// 核心是判断闰年的方法#include <stdio.h> bool isYun (int year) { if (year%4==0 && year%100!=0 || year%400==0) //不是世纪年且能被4整除或者是世纪年能被400整除 return 1; return 0; } int main() { int y, m; int mon[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; while (scanf ("[%d,%d]", &y, &m) != EOF) { if (isYun(y)) //是闰年 mon[1] = 29; else mon[1] = 28; printf ("%d\n", mon[m-1]); getchar(); } return 0; }

03.给出一些标识符,判断合法标识符有多少个

// C语言的合法标识符：以字母或下划线开头，由字母、数字或者下划线构成的字符串#include <stdio.h> #include <ctype.h> #include <string.h> bool isValid (char s[]) { int len = strlen(s); int i; if (isalpha(s[0]) || s[0] == '_') //字母或下划线开头 { for (i = 1; i < len; ++i) { if (isdigit(s[i]) || isalpha(s[i]) || s[i] == '_') continue; else return 0; } return 1; } return 0; } int main() { char id[100]; int tot = 0; while (gets(id)) { if (isValid(id)) { printf("%s\n", id); tot++; } else continue; } printf ("There are %d valid ID\n", tot); return 0; }

04.给出无向图连接矩阵,求各个连同分量.

如：Input:

ABCDEFGH
0 1 0 0 0 0 1 1
1 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0
0 0 0 0 1 1 0 0
0 0 0 1 0 1 0 0
0 0 0 1 1 0 0 0
1 0 0 0 0 0 0 1
0 1 0 0 0 0 1 0

Output:

The 1 th connection is:ABHG
The 2 th connection is:C
The 3 th connection is:DEF

//当对某一顶点进行深搜时，可以遍历到这个顶点所在连通图的所有节点#include <stdio.h> #include <string.h> #define MAXV 50 int graph[MAXV][MAXV]; //图的矩阵表示 bool vis[MAXV]; //是否已访问的标志 char s[MAXV]; //顶点的字母表示 int nVertices; //顶点的个数 void dfs(int v) //深度优先遍历 { vis[v] = 1; //修改为已访问 printf ("%c ", s[v]); int i; for (i = 0; i < nVertices; ++i) if (!vis[i] && graph[v][i]) //没有访问且有边 dfs(i); } int main() { int i, j; while (gets(s)) { memset(vis, 0, sizeof(vis)); //访问标志初始化 memset(graph, 0, sizeof(graph)); //图矩阵初始化 nVertices = strlen(s); for (i = 0; i < nVertices; ++i) //读入图的矩阵 for (j = 0; j < nVertices; ++j) scanf ("%d", &graph[i][j]); int index = 1; for (i = 0; i < nVertices; ++i) //对所有的节点，看是否已经访问 { if (!vis[i]) { printf ("The %dth connection is:", index++); dfs(i); printf ("\n"); } } } return 0; }

5.给出一个整数分解成连续整数的和.

//sum[i] 记录了前i个数的和，从i~j的和用sum[j]-sum[i-1] 计算，若与n相等则输出#include <stdio.h> int main() { int n; int i, j, k; int sum[1000]; //sum[i] --- 前i个数的和 sum[0] = 0; for (i = 1; i < 1000; ++i) sum[i] = sum[i-1] + i; while (scanf ("%d", &n) != EOF) { for (i = 1; i <= n/2; ++i) //i为起始位置，j为终止位置，计算i~j的和 { int ok = 0; for (j = i; j <= n/2; ++j) { if (sum[j] - sum[i-1] == n) //相等 { ok = 1; break; } else if (sum[j] - sum[i-1] > n) //大于，无需将j后移，因为越加会越大 break; } if (ok) { for (k = i; k <= j; k++) //输出 printf ("%d ", k); printf ("\n"); } } } return 0; }

6.给出带括号的四则运算表达式,要求给出逆波兰式

#include <cstdio> #include <cctype> #include <stack> using namespace std; bool compare (char op1, char op2) //比较优先级，op1为当前操作符，op2为栈顶操作符 { // 若op1优先级比op2高，返回1；否则返回0. if (op1 == '+' || op1 == '-') if (op2 == '#' || op2 == '(') return 1; else return 0; else if (op1 == '(') return 1; else if (op2 == '*' || op2 == '/') return 0; else return 1; } int main() { int val; char ch, op; char exp[100]; stack<char> opStack; opStack.push('#'); int i, len; while (gets(exp)) { i = 0; len = strlen(exp); while (i < len) { ch = exp[i]; if (isdigit(ch)) //是数字 { val = ch - '0'; while (isdigit(exp[++i])) val = val * 10 + exp[i] - '0'; i--; //退回 printf ("%d", val); } else if (ch != ')') //不为')‘ { op = opStack.top(); if (compare(ch, op)) //ch > op,压入 opStack.push(ch); else { while (!compare(ch, op)) //ch < op，弹出直到op比ch优先级高 { printf ("%c", op); opStack.pop(); op = opStack.top(); } opStack.push(ch); } } else if (ch == ')') //右括号，弹出栈中操作符直道遇到’（‘，并且将'('弹出 { op = opStack.top(); while (op != '(') { printf ("%c", op); opStack.pop(); op = opStack.top(); } opStack.pop(); } i++; } op = opStack.top(); while (op != '#') { printf ("%c", op); opStack.pop(); op = opStack.top(); } printf("\n"); } return 0; }

7.递归列出的所有选择方法.

例如m=3,n=4时（4选3）,结果为

1,2,3

1,2,4

1,3,4

2,3,4

//求出1~n长度为m的子集#include <stdio.h> void search(int m, int A[], int n, int cur) { int i, j; if (cur == m) //长度为m 输出 { printf ("%d", A[0]); for (i = 1; i < m; ++i) printf (",%d", A[i]); printf ("\n"); return ; } int s = (cur == 0 ? 1 : A[cur-1] + 1); // for (i = s; i <= n; ++i) { A[cur] = i; int ok = 1; for (j = 0; j < cur; ++j) if (A[j] == i) //如果已经能够使用到则返回 { ok = 0; break; } if (ok) search(m, A, n, cur+1); } } int main() { int A[10]; int m, n; while (scanf ("%d %d", &m, &n) != EOF) { search(n, A, m, 0); } return 0; }
代码有可能有BUG，欢迎指正~~