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  • POJ 3579

    Median
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 3528   Accepted: 1001

    Description

    Given N numbers, X1, X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi - Xj∣ (1 ≤ i j N). We can get C(N,2) differences through this work, and now your task is to find the median of the differences as quickly as you can!

    Note in this problem, the median is defined as the (m/2)-th  smallest number if m,the amount of the differences, is even. For example, you have to find the third smallest one in the case of m = 6.

    Input

    The input consists of several test cases.
    In each test case, N will be given in the first line. Then N numbers are given, representing X1, X2, ... , XN, ( Xi ≤ 1,000,000,000  3 ≤ N ≤ 1,00,000 )

    Output

    For each test case, output the median in a separate line.

    Sample Input

    4
    1 3 2 4
    3
    1 10 2
    

    Sample Output

    1
    8

    Source

     
    二分中位数
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 
     6 using namespace std;
     7 
     8 #define maxn 100005
     9 #define INF 200005
    10 typedef long long ll;
    11 
    12 int n;
    13 int a[maxn],dis[maxn];
    14 ll num;
    15 
    16 bool judge(int x) {
    17 
    18     int pos,i = 1,now = 0;
    19     ll sum = 0;
    20     while(i < n) {
    21             pos = upper_bound(dis + 1,dis + n + 1,x + now) - dis;
    22             sum += n - (pos);
    23             now = dis[i];
    24             if((n - 1) - pos + 1 == 0) break;
    25             ++i;
    26     }
    27 
    28     //printf("x = %d sum = %lld
    ",x,sum);
    29 
    30     return num - sum - (num % 2) >= sum;
    31 }
    32 
    33 void solve() {
    34 
    35     num = (ll)n * (n - 1) / 2;
    36 
    37     int l = INF,r = a[n] - a[1];
    38     for(int i = 1; i < n; ++i) {
    39             dis[i] = a[i + 1] - a[1];
    40             l = min(l,a[i + 1] - a[i]);
    41            // printf("%d ",dis[i]);
    42     }
    43     dis[n] = INF;
    44 
    45     //printf("l = %d r = %d
    ",l,r);
    46 
    47     while(l < r) {
    48             int mid = (l + r) >> 1;
    49             if(judge(mid)) {
    50                     r = mid;
    51             } else {
    52                     l = mid + 1;
    53             }
    54     }
    55 
    56     printf("%d
    ",l);
    57 
    58 }
    59 
    60 int main()
    61 {
    62    // freopen("sw.in","r",stdin);
    63 
    64     while(~scanf("%d",&n)) {
    65             for(int i =  1; i <= n; ++i) {
    66                     scanf("%d",&a[i]);
    67             }
    68 
    69             sort(a + 1,a + n + 1);
    70 
    71             solve();
    72     }
    73 
    74 
    75     return 0;
    76 }
    View Code
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  • 原文地址:https://www.cnblogs.com/hyxsolitude/p/3600305.html
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