标签：strong through 暴力 sub output 例题 under 方法 lin
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题目链接：http://poj.org/problem?id=3579
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
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<algorithm> 5 #include<cmath> 6 7 const int MAXN = 100005; 8 9 inline void read(long long &x) 10 { 11 char ch = getchar(),c = ch;x = 0; 12 while(ch < ‘0‘ || ch > ‘9‘) c = ch,ch = getchar(); 13 while(ch <= ‘9‘ && ch >= ‘0‘) x = (x<<1)+(x<<3)+ch-‘0‘,ch = getchar(); 14 if(c == ‘-‘) x = -x; 15 } 16 17 long long n,l,r,mid,base,ans; 18 long long num[MAXN]; 19 20 bool jud(int x) 21 { 22 long long cnt = 0,L = 1,R = 2; 23 for(;L < n;++ L) 24 { 25 while(num[R] - num[L] <= x && R <= n) ++ R; 26 cnt += R-L-1; 27 } 28 if(cnt >= base) return true; 29 return false; 30 } 31 32 int main() 33 { 34 // freopen("1.txt","r",stdin); 35 while(scanf("%d",&n) != EOF) 36 { 37 base = (long long)(n*(n-1)/2+1)/2; 38 for(int i = 1;i <= n;++ i) 39 read(num[i]); 40 std::sort(num+1,num+1+n); 41 l = 1,r = num[n]; 42 while(l <= r) 43 { 44 mid = (l+r)>>1; 45 if(jud(mid)) ans = mid,r = mid-1; 46 else l = mid+1; 47 } 48 printf("%lld\n",ans); 49 } 50 return 0; 51 }
标签：strong through 暴力 sub output 例题 under 方法 lin
原文地址：http://www.cnblogs.com/shingen/p/7643852.html
