国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

1029. Median (25)

2019-11-14 08:49:29
字體:
來源:轉載
供稿:網友

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 median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.

Given two increasing sequences of integers, you are asked to find their median.

Input

Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (<=1000000) is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.

Output

For each test case you should output the median of the two given sequences in a line.

Sample Input 4 11 12 13 14 5 9 10 15 16 17 Sample Output 13

#include<cstdio>const int INF=0x7fffffff;const int maxn=1000010;int a[maxn],b[maxn];int main(){ int n1,n2; scanf("%d",&n1); for(int i=0;i<n1;i++){ scanf("%d",&a[i]); } scanf("%d",&n2); for(int i=0;i<n1;i++){ scanf("%d",&b[i]); } a[n1]=b[n2]=INF;//防止在掃描過程中,其中一個序列已掃描完,但還沒到中位數的情況 int pos=(n1+n2-1)/2; int i=0,j=0,count=0; while(count<pos){ if(a[i]<b[j]) i++; else j++; count++; } if(a[i]<b[j])
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 临桂县| 九寨沟县| 房山区| 瑞丽市| 察雅县| 分宜县| 平武县| 清徐县| 綦江县| 京山县| 宣威市| 金秀| 河东区| 陇南市| 邻水| 兰西县| 宁城县| 康马县| 公主岭市| 阿拉尔市| 綦江县| 景洪市| 河西区| 昌图县| 年辖:市辖区| 榆林市| 长顺县| 黄陵县| 娱乐| 兰坪| 来凤县| 浦城县| 阿城市| 禄丰县| 黑山县| 湘阴县| 永靖县| 蕉岭县| 镇安县| 泾阳县| 平果县|