Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the medi…

1029. Median (25) 时间限制 1000 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median…

1029 Median (25 分)   Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of two sequen…

题意: 输入一个正整数N(<=2e5),接着输入N个非递减序的长整数. 输入一个正整数N(<=2e5),接着输入N个非递减序的长整数.(重复一次) 输出两组数合并后的中位数.(200ms,合并后排序会超时,利用两组数是有序的进行模拟) AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; int main(){ int n; cin>>…

1032 Sharing (25)(25 分) To store English words, one method is to use linked lists and store a word letter by letter. To save some space, we may let the words share the same sublist if they share the same suffix. For example, "loading" and "…

题目 Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the m…

1029 Median (25 分)   Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of two sequen…

题目 Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of two sequences is defined to…

这道题算有点难,心目中理想的难度. 不能前怕狼后怕虎,一会担心超时,一会又担心内存过大,直接撸 将三部分分别保存到vector 有意思的在于输出 分别输出第一个的add和num 中间输出nextadd ,换行,add,num 最后输出尾元素的next为-1 #include<iostream> #include<stdio.h> #include<map> #include<string> #include<algorithm> #include…

[题意] 给两个有序数组,寻找两个数组组成后的中位数,要求时间复杂度为O(log(n+m)). [题解] 感觉这道题想法非常妙!! 假定原数组为a,b,数组长度为lena,lenb. 那么中位数一定是第k = (lena + lenb + 1)/ 2小的数,如果是数组长度和是偶数的话就是第k = (lena + lenb + 1)/ 2小和第k = (lena + lenb + 2)/ 2小的数,所以我们可以把问题转化为求第k小的数. 然后分别对a,b找第k / 2小的数,假如a[k / 2 -…