求第k大的数

【问题描述】 求n个数中第k大的数
【输入形式】 第一行n k，第二行为n个数，都以空格分开
【输出形式】 第k大的数
【样例输入】
10 3
18 21 11 26 12 2 9 33 43 28

【样例输出】

28

【评分标准】时间复杂度大于等于O(kn)的方法得一半分，时间复杂度小于等于O(nlog2k)得满分。

【提示】

1. 分析各种排序或查找算法的优缺点，分析解决具体问题的时间复杂度，进而找出更高效的算法。

2. n与k的值不同，不同算法的效率也会有影响，如n=10, k=9时，可以找第2小的数。

#include<iostream>
using namespace std;int QSort(int a[], int left, int right, int rk)
{int low = left;int high = right;int flag = a[low];while(low < high){while(a[high] <= flag && low < high){high --;}a[low] = a[high];while(a[low] >= flag && low < high){low ++;}a[high] = a[low];}a[low] = flag;if(low == rk - 1)return a[low];else if(low > rk - 1)return QSort(a, left, low - 1, rk);elsereturn QSort(a, low + 1, right, rk - low);
}
int main()
{int n, k;cin>>n>>k;int i = n;int j = -1;int a[n];while(i --){cin>>a[++ j];}cout<<QSort(a, 0, n - 1, k);
}

查找3个数组的最小共同元素

【问题描述】查找3个数组的最小共同元素
【输入形式】三个数组，均以0代表输入结束
【输出形式】最小共同元素
【样例输入】

1 3 5 7 8 9 0

2 4 6 8 10 12 14 16 18 0

-1 3 8 16 18 19 20 168 198 0
【样例输出】

8

#include<iostream>
using namespace std;
int a[1000], b[1000], c[1000];
int main()
{int i = 0;int num = 0;for(i = 0; ; i ++){cin>>num;if(num == 0) break;a[i] = num; //i取到数组最后一个下标}int alen = i;for(i = 0; ; i ++){cin>>num;if(num == 0) break;b[i] = num; //i取到数组最后一个下标}int blen = i;for(i = 0; ; i ++){cin>>num;if(num == 0) break;c[i] = num; //i取到数组最后一个下标}int clen = i;int pa = 0, pb = 0, pc = 0;while(pa <= alen && pb <= blen && pc <= clen){if(a[pa] == b[pb] && b[pb] == c[pc]){cout<<a[pa];break;}while(a[pa] < b[pb] && pa <= alen){pa ++;}while(b[pb] < a[pa] && pb <= blen){pb ++;}while(c[pc] < b[pb] && pc <= clen){pc ++;}}return 0;
}

查找一个循环顺序数组的最小元素

【问题描述】以循环排列的一组顺序的数据，存储在一维数组中，查找最小元素并输出。
【输入形式】一组数据，以0结束输入
【输出形式】最小元素
【样例输入】7 9 11 1 3 5 0
【样例输出】1

#include<iostream>
#define N 100
using namespace std;int FindMin(int a[], int low, int high)
{int mid = (low + high) / 2;if(a[low] < a[high])return a[low];else{if(a[low] < a[mid]){return FindMin(a, mid + 1, high);}else if(a[low] == a[mid]) return a[low];else{return FindMin(a, low + 1, mid);}}
}
int main()
{int a[N];int i = 0;int num = 0;for(i = 0; ;i ++){cin>>num;if(num == 0) break;a[i] = num;}cout<<FindMin(a, 0, i - 1); //i取不到return 0;
}

Crazy Search

【题目来源】1200 – Crazy Search (poj.org) 请前往此链接提交检测代码

Description

Many people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a given text. Such number could be the number of different substrings of a given size that exist in the text. As you soon will discover, you really need the help of a computer and a good algorithm to solve such a puzzle.
Your task is to write a program that given the size, N, of the substring, the number of different characters that may occur in the text, NC, and the text itself, determines the number of different substrings of size N that appear in the text.

As an example, consider N=3, NC=4 and the text “daababac”. The different substrings of size 3 that can be found in this text are: “daa”; “aab”; “aba”; “bab”; “bac”. Therefore, the answer should be 5.

Input

The first line of input consists of two numbers, N and NC, separated by exactly one space. This is followed by the text where the search takes place. You may assume that the maximum number of substrings formed by the possible set of characters does not exceed 16 Millions.

Output

The program should output just an integer corresponding to the number of different substrings of size N found in the given text.

Sample Input

3 4
daababac
Sample Output

5

#include<iostream>
#include<string>
#include<string.h>
using namespace std;const int N = 1600000; // 定义16000000为什么不能运行int main()
{int res = 0;string s;int sonlen;int sysnum; //字符串中可能出现的字符种类数cin>>sonlen;cin>>sysnum;cin>>s;int slen = s.length(); //调用string类的类函数int i = 0;bool Hash[N];memset(Hash, 0, sizeof(Hash));for(i = 0; i <= slen - sonlen; i ++){string temp = s.substr(i,3); //截取字符串片段int pos = 0;int j = 0;for(j = 0; j < sonlen; j ++){int k = 1;int t = int(temp[j]);for(k = j + 1; k <= sysnum; k ++){t *= sysnum;}pos += t;}if(!Hash[pos]){Hash[pos] = 1;res ++;}}cout<<res;return 0;
}