liuchuo – 第117页 – 我不管，反正我最萌～

1070. 结绳(25)-PAT乙级真题

给定一段一段的绳子，你需要把它们串成一条绳。每次串连的时候，是把两段绳子对折，再如下图所示套接在一起。这样得到的绳子又被当成是另一段绳子，可以再次对折去跟另一段绳子串连。每次串连后，原来两段绳子的长度就会减半。
给定N段绳子的长度，你需要找出它们能串成的绳子的最大长度。

输入格式：

每个输入包含1个测试用例。每个测试用例第1行给出正整数N (2 <= N <= 10⁴)；第2行给出N个正整数，即原始绳段的长度，数字间以空格分隔。所有整数都不超过10⁴。

输出格式：

在一行中输出能够串成的绳子的最大长度。结果向下取整，即取为不超过最大长度的最近整数。

输入样例：
8
10 15 12 3 4 13 1 15
输出样例：
14

分析：因为所有长度都要串在一起，每次都等于(旧的绳子长度+新的绳子长度)/2，所以越是早加入绳子长度中的段，越要对折的次数多，所以既然希望绳子长度是最长的，就必须让长的段对折次数尽可能的短。所以将所有段从小到大排序，然后从头到尾从小到大分别将每一段依次加入结绳的绳子中，最后得到的结果才会是最长的结果～

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {
    int n;
    scanf("%d", &n);
    vector<int> v(n);
    for (int i = 0; i < n; i++)
        scanf("%d", &v[i]);
    sort(v.begin(), v.end());
    int result = v[0];
    for (int i = 1; i < n; i++)
        result = (result + v[i]) / 2;
    printf("%d", result);
    return 0;
}

#include <iostream>

#include <algorithm>

#include <vector>

using namespace std;

int main() {

int n;

scanf("%d", &n);

vector<int> v(n);

for (int i = 0; i < n; i++)

scanf("%d", &v[i]);

sort(v.begin(), v.end());

int result = v[0];

for (int i = 1; i < n; i++)

result = (result + v[i]) / 2;

printf("%d", result);

return 0;

}

1069. 微博转发抽奖(20)-PAT乙级真题

小明PAT考了满分，高兴之余决定发起微博转发抽奖活动，从转发的网友中按顺序每隔N个人就发出一个红包。请你编写程序帮助他确定中奖名单。

输入格式：

输入第一行给出三个正整数M（<= 1000）、N和S，分别是转发的总量、小明决定的中奖间隔、以及第一位中奖者的序号（编号从1开始）。随后M行，顺序给出转发微博的网友的昵称（不超过20个字符、不包含空格回车的非空字符串）。

注意：可能有人转发多次，但不能中奖多次。所以如果处于当前中奖位置的网友已经中过奖，则跳过他顺次取下一位。

输出格式：

按照输入的顺序输出中奖名单，每个昵称占一行。如果没有人中奖，则输出“Keep going…”。

输入样例1：
9 3 2
Imgonnawin!
PickMe
PickMeMeMeee
LookHere
Imgonnawin!
TryAgainAgain
TryAgainAgain
Imgonnawin!
TryAgainAgain
输出样例1：
PickMe
Imgonnawin!
TryAgainAgain
输入样例2：
2 3 5
Imgonnawin!
PickMe
输出样例2：
Keep going…

分析：用mapp存储当前用户有没有已经中奖过～当输入的时候，判断当前字符串是否已经在mapp中出现过，如果出现过就将s+1。每次判断i是否等于s，如果等于s且当前用户没有中过奖，就将它的名字输出，并且s = s + n～并将mapp[str]标记为1，且flag标记为true表示有过人中奖。最后flag如果依然是false说明要输出Keep going…

#include <iostream>
#include <map>
using namespace std;
int main() {
    int m, n, s;
    scanf("%d%d%d", &m, &n, &s);
    string str;
    map<string, int> mapp;
    bool flag = false;
    for (int i = 1; i <= m; i++) {
        cin >> str;
        if (mapp[str] == 1) s = s + 1;
        if (i == s && mapp[str] == 0) {
            mapp[str] = 1;
            cout << str << endl;
            flag = true;
            s = s + n;
        }
    }
    if (flag == false) cout << "Keep going...";
    return 0;
}

#include <iostream>

#include <map>

using namespace std;

int main() {

int m, n, s;

scanf("%d%d%d", &m, &n, &s);

string str;

map<string, int> mapp;

bool flag = false;

for (int i = 1; i <= m; i++) {

cin >> str;

if (mapp[str] == 1) s = s + 1;

if (i == s && mapp[str] == 0) {

mapp[str] = 1;

cout << str << endl;

flag = true;

s = s + n;

}

if (flag == false) cout << "Keep going...";

return 0;

}

PAT 1125. Chain the Ropes (25)-甲级

Given some segments of rope, you are supposed to chain them into one rope. Each time you may only fold two segments into loops and chain them into one piece, as shown by the figure. The resulting chain will be treated as another segment of rope and can be folded again. After each chaining, the lengths of the original two segments will be halved.

Your job is to make the longest possible rope out of N given segments.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (2 <= N <= 104). Then N positive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 104.

Output Specification:

For each case, print in a line the length of the longest possible rope that can be made by the given segments. The result must be rounded to the nearest integer that is no greater than the maximum length.

Sample Input:
8
10 15 12 3 4 13 1 15
Sample Output:
14

题目大意：给定一段一段的绳子，你需要把它们串成一条绳。每次串连的时候，是把两段绳子对折，再如下图所示套接在一起。这样得到的绳子又被当成是另一段绳子，可以再次对折去跟另一段绳子串连。每次串连后，原来两段绳子的长度就会减半。给定N段绳子的长度，你需要找出它们能串成的绳子的最大长度~
分析：因为所有长度都要串在一起，每次都等于(旧的绳子长度+新的绳子长度)/2，所以越是早加入绳子长度中的段，越要对折的次数多，所以既然希望绳子长度是最长的，就必须让长的段对折次数尽可能的短。所以将所有段从小到大排序，然后从头到尾从小到大分别将每一段依次加入结绳的绳子中，最后得到的结果才会是最长的结果～

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {
    int n;
    scanf("%d", &n);
    vector<int> v(n);
    for (int i = 0; i < n; i++)
        scanf("%d", &v[i]);
    sort(v.begin(), v.end());
    int result = v[0];
    for (int i = 1; i < n; i++)
        result = (result + v[i]) / 2;
    printf("%d", result);
    return 0;
}

#include <iostream>

#include <algorithm>

#include <vector>

using namespace std;

int main() {

int n;

scanf("%d", &n);

vector<int> v(n);

for (int i = 0; i < n; i++)

scanf("%d", &v[i]);

sort(v.begin(), v.end());

int result = v[0];

for (int i = 1; i < n; i++)

result = (result + v[i]) / 2;

printf("%d", result);

return 0;

}

1124. Raffle for Weibo Followers (20)-PAT甲级真题

John got a full mark on PAT. He was so happy that he decided to hold a raffle（抽奖） for his followers on Weibo — that is, he would select winners from every N followers who forwarded his post, and give away gifts. Now you are supposed to help him generate the list of winners.

Input Specification:

Each input file contains one test case. For each case, the first line gives three positive integers M (<= 1000), N and S, being the total number of forwards, the skip number of winners, and the index of the first winner (the indices start from 1). Then M lines follow, each gives the nickname (a nonempty string of no more than 20 characters, with no white space or return) of a follower who has forwarded John’s post.

Note: it is possible that someone would forward more than once, but no one can win more than once. Hence if the current candidate of a winner has won before, we must skip him/her and consider the next one.

Output Specification:

For each case, print the list of winners in the same order as in the input, each nickname occupies a line. If there is no winner yet, print “Keep going…” instead.

Sample Input 1:
9 3 2
Imgonnawin!
PickMe
PickMeMeMeee
LookHere
Imgonnawin!
TryAgainAgain
TryAgainAgain
Imgonnawin!
TryAgainAgain
Sample Output 1:
PickMe
Imgonnawin!
TryAgainAgain
Sample Input 2:
2 3 5
Imgonnawin!
PickMe
Sample Output 2:
Keep going…

题目大意：小明PAT考了满分，高兴之余决定发起微博转发抽奖活动，从转发的网友中按顺序每隔N个人就发出一个红包。请你编写程序帮助他确定中奖名单。注意：可能有人转发多次，但不能中奖多次。所以如果处于当前中奖位置的网友已经中过奖，则跳过他顺次取下一位。按照输入的顺序输出中奖名单，每个昵称占一行。如果没有人中奖，则输出“Keep going…”
分析：用mapp存储当前用户有没有已经中奖过～当输入的时候，判断当前字符串是否已经在mapp中出现过，如果出现过就将s+1。每次判断i是否等于s，如果等于s且当前用户没有中过奖，就将它的名字输出，并且s = s + n～并将mapp[str]标记为1，且flag标记为true表示有过人中奖。最后flag如果依然是false说明要输出Keep going…

#include <iostream>
#include <map>
using namespace std;
int main() {
    int m, n, s;
    scanf("%d%d%d", &m, &n, &s);
    string str;
    map<string, int> mapp;
    bool flag = false;
    for (int i = 1; i <= m; i++) {
        cin >> str;
        if (mapp[str] == 1) s = s + 1;
        if (i == s && mapp[str] == 0) {
            mapp[str] = 1;
            cout << str << endl;
            flag = true;
            s = s + n;
        }
    }
    if (flag == false) cout << "Keep going...";
    return 0;
}

#include <iostream>

#include <map>

using namespace std;

int main() {

int m, n, s;

scanf("%d%d%d", &m, &n, &s);

string str;

map<string, int> mapp;

bool flag = false;

for (int i = 1; i <= m; i++) {

cin >> str;

if (mapp[str] == 1) s = s + 1;

if (i == s && mapp[str] == 0) {

mapp[str] = 1;

cout << str << endl;

flag = true;

s = s + n;

}

if (flag == false) cout << "Keep going...";

return 0;

}

使用Dev-C++查看vector数组中的变量值

可以通过调试的时候添加查看：比如说有一个长度为3的vector v，如果想要查看v[0]的值，就在添加查看中写 *(&v[0])

如果想要查看整个数组的值，就可以写*(&v[0])@3

@后面的数字表示想要查看的长度，这里vector的长度是3所以可以写3就能看到所有的值～

LeetCode 524. Longest Word in Dictionary through Deleting

Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.

Example 1:
Input:
s = “abpcplea”, d = [“ale”,”apple”,”monkey”,”plea”]

Output:
“apple”
Example 2:
Input:
s = “abpcplea”, d = [“a”,”b”,”c”]

Output:
“a”
Note:
All the strings in the input will only contain lower-case letters.
The size of the dictionary won’t exceed 1,000.
The length of all the strings in the input won’t exceed 1,000.

题目大意：给一个string s和一个string字典d，找字典中的某个string，寻找s的子串（满足可以通过删除s中某些元素后得到该string），寻找满足条件的字符串中最长的一个，如果有多个长度相等的就返回字典序中最小的那个，如果一个都没有满足条件的string，就返回一个空字符串～
分析：遍历字典中的某一个字符串，设当前字符串的下标为index，对于当前字符串d[index]，使用两个指针i和j分别从头到尾遍历s和d[index]，随着i指针的增加，如果j指针所指元素和i指针所指元素相同就向后移动一位，当i指针都指完的时候，j如果也指完了说明满足条件，当前d[index]是s的子串，如果当前d[index]的长度比保存的result字符串长度长，就更新result，或者一样长但是字典序排列中d[index]比result小，也要更新result，最后返回result～

class Solution {
public:
    string findLongestWord(string s, vector<string>& d) {
        string result = "";
        for (int index = 0; index < d.size(); index++) {
            int i, j;
            for (i = 0, j = 0; i < s.length() && j < d[index].length(); i++) {
                if (s[i] == d[index][j]) j++;
            }
            if (j == d[index].length() && (result.length() < d[index].length() || (result.length() == d[index].length() && result > d[index])))
                result = d[index];
        }
        return result;
    }
};

class Solution {

public:

string findLongestWord(string s, vector<string>& d) {

string result = "";

for (int index = 0; index < d.size(); index++) {

int i, j;

for (i = 0, j = 0; i < s.length() && j < d[index].length(); i++) {

if (s[i] == d[index][j]) j++;

}

if (j == d[index].length() && (result.length() < d[index].length() || (result.length() == d[index].length() && result > d[index])))

result = d[index];

}

return result;

}

};