7-1 Good in C (20分)
When your interviewer asks you to write "Hello World" using C, can you do as the following figure shows?
Input Specification:
Each input file contains one test case. For each case, the first part gives the 26 capital English letters A-Z, each in a 7×5 matrix of C's and .'s. Then a sentence is given in a line, ended by a return. The sentence is formed by several words (no more than 10 continuous capital English letters each), and the words are separated by any characters other than capital English letters.
It is guaranteed that there is at least one word given.
Output Specification:
For each word, print the matrix form of each of its letters in a line, and the letters must be separated by exactly one column of space. There must be no extra space at the beginning or the end of the word.
Between two adjacent words, there must be a single empty line to separate them. There must be no extra line at the beginning or the end of the output.
Sample Input:
..C..
.C.C.
C...C
CCCCC
C...C
C...C
C...C
CCCC.
C...C
C...C
CCCC.
C...C
C...C
CCCC.
.CCC.
C...C
C....
C....
C....
C...C
.CCC.
CCCC.
C...C
C...C
C...C
C...C
C...C
CCCC.
CCCCC
C....
C....
CCCC.
C....
C....
CCCCC
CCCCC
C....
C....
CCCC.
C....
C....
C....
CCCC.
C...C
C....
C.CCC
C...C
C...C
CCCC.
C...C
C...C
C...C
CCCCC
C...C
C...C
C...C
CCCCC
..C..
..C..
..C..
..C..
..C..
CCCCC
CCCCC
....C
....C
....C
....C
C...C
.CCC.
C...C
C..C.
C.C..
CC...
C.C..
C..C.
C...C
C....
C....
C....
C....
C....
C....
CCCCC
C...C
C...C
CC.CC
C.C.C
C...C
C...C
C...C
C...C
C...C
CC..C
C.C.C
C..CC
C...C
C...C
.CCC.
C...C
C...C
C...C
C...C
C...C
.CCC.
CCCC.
C...C
C...C
CCCC.
C....
C....
C....
.CCC.
C...C
C...C
C...C
C.C.C
C..CC
.CCC.
CCCC.
C...C
CCCC.
CC...
C.C..
C..C.
C...C
.CCC.
C...C
C....
.CCC.
....C
C...C
.CCC.
CCCCC
..C..
..C..
..C..
..C..
..C..
..C..
C...C
C...C
C...C
C...C
C...C
C...C
.CCC.
C...C
C...C
C...C
C...C
C...C
.C.C.
..C..
C...C
C...C
C...C
C.C.C
CC.CC
C...C
C...C
C...C
C...C
.C.C.
..C..
.C.C.
C...C
C...C
C...C
C...C
.C.C.
..C..
..C..
..C..
..C..
CCCCC
....C
...C.
..C..
.C...
C....
CCCCC
HELLO~WORLD!Sample Output:
C...C CCCCC C.... C.... .CCC.
C...C C.... C.... C.... C...C
C...C C.... C.... C.... C...C
CCCCC CCCC. C.... C.... C...C
C...C C.... C.... C.... C...C
C...C C.... C.... C.... C...C
C...C CCCCC CCCCC CCCCC .CCC.
C...C .CCC. CCCC. C.... CCCC.
C...C C...C C...C C.... C...C
C...C C...C CCCC. C.... C...C
C.C.C C...C CC... C.... C...C
CC.CC C...C C.C.. C.... C...C
C...C C...C C..C. C.... C...C
C...C .CCC. C...C CCCCC CCCC.思路分析：本题有很多细节需要处理。首先第一个字符不一定是大写字母，故要先找到第一个大写字母。其次，中间的间隔字符可能是空格，故要整行读取（一旦使用了整行读取函数，前面的输入后面记得添加getchar()来吸收换行符）。最后，末尾不一定有其他字符，故可以考虑人为加上一个非大写字母的字符，从而便于处理，这种操作之前也是见过的。
再来看图形如何输出。由于输出只能一行一行输出，故不能每个单词一个字母一个字母输出，而要每个单词的每个字母的相同行先输出，然后再输出下一行。这其中要想清楚，可以添加中文注释防止搞混。
参考代码：
#include
using namespace std;
int main(){
string letter[30][10];
for(int i = 0; i < 26; i ++){
for(int j = 0; j < 7; j++){
cin >> letter[i][j];
getchar();
}
}
string seq;
int pos, flag;
getline(cin, seq);
seq += "^";
int i = 0;
while(seq[i] > 'Z' || seq[i] < 'A') i++;
pos = flag = i;
for(; i < seq.length()+1; i++){
if(seq[i] > 'Z' || seq[i] < 'A'){
if(pos != flag) printf("\n");
for(int j = 0; j < 7; j++){//每个字母的每一行
for(int k = pos; k < i; k++){
cout << letter[seq[k]-'A'][j]; //不同字母
printf("%s", k == i-1 ? "\n" : " ");
}
}
while(i+1 < seq.length() && seq[i+1] > 'Z' || seq[i+1] < 'A') i++;
pos = i+1;
}
}
return 0;
}
7-2 Block Reversing (25分)
Given a singly linked list L. Let us consider every K nodes as a block (if there are less than K nodes at the end of the list, the rest of the nodes are still considered as a block). Your job is to reverse all the blocks in L. For example, given L as 1→2→3→4→5→6→7→8 and K as 3, your output must be 7→8→4→5→6→1→2→3.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10^5) which is the total number of nodes, and a positive K (≤N) which is the size of a block. The address of a node is a 5-digit nonnegative integer, and NULL is represented by −1.
Then N lines follow, each describes a node in the format:
Address Data Nextwhere Address is the position of the node, Data is an integer, and Next is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218Sample Output:
71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1思路分析：本题属于链表的常规题，静态链表解决即可，主要注意每一块需要使用一个向量数组来存储，是二维的。
可能存在无效结点，故从头结点开始向后访问直到尾结点，并存入一个向量中。然后每k个结点加入一个向量中，并存入开辟的数组中。由于最后可能不满k个，链表就结束了，故最后若临时向量非空，则加入结果数组中即可。然后顺次输出，注意分三种情况。
参考代码：
#include
using namespace std;
struct node{
int Address, Data, Next;
}link[100010];
vector block[100010];
vector temp;
int main(){
int head, node_num, block_size;
int add, dat, nex;
scanf("%d %d %d", &head, &node_num, &block_size);
for(int i = 0; i < node_num; i++){
scanf("%d %d %d", &add, &dat, &nex);
link[add].Address = add;
link[add].Data = dat;
link[add].Next = nex;
}
int cot = 0;
int block_num = 0;
while(head != -1){
temp.push_back(link[head]);
cot++;
if(cot % block_size == 0){
block[block_num++] = temp;
temp.clear();
}
head = link[head].Next;
}
if(!temp.empty()) block[block_num++] = temp;
reverse(block, block+block_num);
for(int i = 0; i < block_num; i++){
for(int j = 0; j < block[i].size(); j++){
if(j == block[i].size()-1){
if(i == block_num-1) printf("%05d %d -1\n", block[i][j].Address, block[i][j].Data);
else printf("%05d %d %05d\n", block[i][j].Address, block[i][j].Data, block[i+1][0].Address);
}
else printf("%05d %d %05d\n", block[i][j].Address, block[i][j].Data, block[i][j+1].Address);
}
}
return 0;
}
7-3 Summit (25分)
A summit (峰会) is a meeting of heads of state or government. Arranging the rest areas for the summit is not a simple job. The ideal arrangement of one area is to invite those heads so that everyone is a direct friend of everyone.
Now given a set of tentative arrangements, your job is to tell the organizers whether or not each area is all set.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N (≤ 200), the number of heads in the summit, and M, the number of friendship relations. Then M lines follow, each gives a pair of indices of the heads who are friends to each other. The heads are indexed from 1 to N.
Then there is another positive integer K (≤ 100), and K lines of tentative arrangement of rest areas follow, each first gives a positive number L (≤N), then followed by a sequence of L distinct indices of the heads. All the numbers in a line are separated by a space.
Output Specification:
For each of the K areas, print in a line your advice in the following format:
- if in this area everyone is a direct friend of everyone, and no friend is missing (that is, no one else is a direct friend of everyone in this area), print
Area X is OK.. - if in this area everyone is a direct friend of everyone, yet there are some other heads who may also be invited without breaking the ideal arrangement, print
Area X may invite more people, such as H.whereHis the smallest index of the head who may be invited. - if in this area the arrangement is not an ideal one, then print
Area X needs help.so the host can provide some special service to help the heads get to know each other.
Here X is the index of an area, starting from 1 to K.
Sample Input:
8 10
5 6
7 8
6 4
3 6
4 5
2 3
8 2
2 7
5 3
3 4
6
4 5 4 3 6
3 2 8 7
2 2 3
1 1
2 4 6
3 3 2 1Sample Output:
Area 1 is OK.
Area 2 is OK.
Area 3 is OK.
Area 4 is OK.
Area 5 may invite more people, such as 3.
Area 6 needs help.思路分析：这种题最近很常见，但牢记不要用图的遍历或并查集啥的，逐个检验即可。本题即是先判一房间人是否互相都认识，再看是否存在一个人与该房间每个人都认识即可。
参考代码：
#include
using namespace std;
int rela[210][210];
int main(){
int h1, h2;
int head_num, rela_num;
scanf("%d %d", &head_num, &rela_num);
for(int i = 0; i < rela_num; i++){
scanf("%d %d", &h1, &h2);
rela[h1][h2] = rela[h2][h1] = 1;
}
int rest_num, in_num;
scanf("%d", &rest_num);
for(int i = 1; i <= rest_num; i++){
int exist[210] = {0};
int flag = 1;
int in[210];
scanf("%d", &in_num);
for(int j = 0; j < in_num; j++){
scanf("%d", &in[j]);
exist[in[j]] = 1;
if(flag){
for(int k = j-1; k >= 0; k--){
if(!rela[in[j]][in[k]]) flag = 0;
}
}
}
if(!flag) printf("Area %d needs help.\n", i);
else{
int more = 0;
int j = 1;
for(; j <= head_num; j++){
if(!exist[j]){//不在该房间的人
int k;
for(k = 0; k < in_num; k++){//与该房间每个人是否认识
if(!rela[j][in[k]]) break;
}
if(k == in_num){
more = 1;
break;
}
}
}
if(more) printf("Area %d may invite more people, such as %d.\n", i, j);
else printf("Area %d is OK.\n", i);
}
}
return 0;
}
7-4 Cartesian Tree (30分)
A Cartesian tree is a binary tree constructed from a sequence of distinct numbers. The tree is heap-ordered, and an inorder traversal returns the original sequence. For example, given the sequence { 8, 15, 3, 4, 1, 5, 12, 10, 18, 6 }, the min-heap Cartesian tree is shown by the figure.
Your job is to output the level-order traversal sequence of the min-heap Cartesian tree.
Input Specification:
Each input file contains one test case. Each case starts from giving a positive integer N (≤30), and then N distinct numbers in the next line, separated by a space. All the numbers are in the range of int.
Output Specification:
For each test case, print in a line the level-order traversal sequence of the min-heap Cartesian tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the beginning or the end of the line.
Sample Input:
10
8 15 3 4 1 5 12 10 18 6Sample Output:
1 3 5 8 4 6 15 10 12 18思路分析：最小数是根，由于中序序列和原始序列一样，故左子树的全部结点在序列中根的左边，右子树同理，然后递归建树即可。
参考代码：
#include
using namespace std;
struct node{
int data;
node *left, *right;
};
vector seq;
node* build(int l, int r){
if(l > r) return NULL;
node* root = new node;
int min_num = seq[l], min_pos = l;
for(int i = l+1; i <= r; i++){
if(seq[i] < min_num){
min_num = seq[i];
min_pos = i;
}
}
root->data = min_num;
root->left = build(l, min_pos-1);
root->right = build(min_pos+1, r);
return root;
}
void BFS(node* root){
queue q;
q.push(root);
while(!q.empty()){
node* top = q.front();
q.pop();
if(top->left != NULL) q.push(top->left);
if(top->right != NULL) q.push(top->right);
printf("%d%s", top->data, q.empty() ? "\n" : " ");
}
}
int main(){
int n;
scanf("%d", &n);
seq.resize(n);
for(int i = 0; i < n; i++) scanf("%d", &seq[i]);
node* root = new node;
root = build(0, n-1);
BFS(root);
return 0;
}
第一次模拟考试获得了满分，还是挺开心的，不过题目也不算太难。