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题目链接:http://acm.swust.edu.cn/problem/1094/
中位数(又称中值,英语:Median),统计学中的专有名词,代表一个样本、种群或概率分布中的一个数值,其可将数值集合划分为相等的上下两部分。对于有限的数集,可以通过把所有观察值高低排序后找出正中间的一个作为中位数。如果观察值有偶数个,则中位数不唯一,通常取最中间的两个数值的平均数作为中位数。
多组输入
第一行:一个正整数N (0<N<1000000)
第二行:N个正整数。(0=<A[i]<2^30)
每组数据先输出”Case X:”,X表示测试数据的编号,从1开始。
第二行输出N个数,第i个数对应数组前i个值的中位数。(精确到小数点后一位)
1
2
3
4
5
|
5
1 2 3 4 5
6
2 5 4 8 7 4
|
1
2
3
4
5
|
Case 1:
1.0 1.5 2.0 2.5 3.0
Case 2:
2.0 3.5 4.0 4.5 5.0 4.5
|
1 #include <iostream> 2 #include <set> 3 #include <cstdio> 4 using namespace std; 5 6 #define rep(i,a,b) for(int i=a;i<=b;i++) 7 int t, n; 8 multiset<double>mpt; 9 10 void mid_num(){ 11 double x; 12 printf("Case %d:\r\n", ++t); 13 mpt.clear(); 14 scanf("%lf", &x); 15 mpt.insert(x); 16 multiset <double> ::iterator it = mpt.begin(); 17 printf("%.1lf", x); 18 rep(i, 2, n){ 19 scanf("%lf", &x); 20 mpt.insert(x); 21 if (x < *it){ 22 if (i & 1) printf(" %.1lf", *(--it)); 23 else { 24 multiset <double> ::iterator it1 = it; 25 printf(" %.1lf", (*it + *(--it1)) / 2.0); 26 } 27 } 28 else { 29 if (i & 1) printf(" %.1lf", *it); 30 else { 31 ++it; 32 multiset <double> ::iterator it1 = it; 33 printf(" %.1lf", (*it + *(--it1)) / 2.0); 34 35 } 36 } 37 } 38 printf("\r\n"); 39 } 40 41 int main(){ 42 while (~scanf("%d", &n)) 43 mid_num(); 44 return 0; 45 }
堆排序的代码也贴出来吧(原谅我太懒没有优化这代码~~~)
1 #include <stdio.h> 2 #include <string.h> 3 int lstack[500001], ltot, rstack[500001], rtot, mid; 4 int Max(int a, int b){ 5 return a > b ? a : b; 6 } 7 int Min(int a, int b){ 8 return a<b ? a : b; 9 } 10 void lup(int step){ 11 while (step != 1){ 12 if (lstack[step]>lstack[step / 2])lstack[step] ^= lstack[step / 2] ^= lstack[step] ^= lstack[step / 2]; 13 else break; 14 step = step / 2; 15 } 16 if (step == 1 && lstack[1] > mid) lstack[1] ^= mid ^= lstack[1] ^= mid; 17 } 18 void ldown(){ 19 if (ltot > 1 && mid < lstack[1]) mid ^= lstack[1] ^= mid ^= lstack[1]; 20 else return; 21 int step = 1; 22 while (step * 2 < ltot){ 23 if (step * 2 + 1 >= ltot){ 24 if (lstack[step] < lstack[step * 2]) 25 lstack[step] ^= lstack[step * 2] ^= lstack[step] ^= lstack[step * 2], step = step * 2; 26 else return; 27 } 28 else{ 29 if (lstack[step] <= lstack[step * 2] && lstack[step * 2 + 1] <= lstack[step * 2]) 30 lstack[step] ^= lstack[step * 2] ^= lstack[step] ^= lstack[step * 2], step = step * 2; 31 else if (lstack[step] <= lstack[step * 2 + 1] && lstack[step * 2] <= lstack[step * 2 + 1]) 32 lstack[step] ^= lstack[step * 2 + 1] ^= lstack[step] ^= lstack[step * 2 + 1], step = step * 2 + 1; 33 else return; 34 } 35 } 36 } 37 void rup(int step){ 38 while (step != 1){ 39 if (rstack[step]<rstack[step / 2])rstack[step] ^= rstack[step / 2] ^= rstack[step] ^= rstack[step / 2]; 40 else break; 41 step = step / 2; 42 } 43 if (step == 1 && rstack[1]<mid) rstack[1] ^= mid ^= rstack[1] ^= mid; 44 } 45 void rdown(){ 46 if (rtot>1 && mid>rstack[1]) mid ^= rstack[1] ^= mid ^= rstack[1]; 47 else return; 48 int step = 1; 49 while (step * 2 < rtot){ 50 if (step * 2 + 1 >= rtot){ 51 if (rstack[step] > rstack[step * 2]) 52 rstack[step] ^= rstack[step * 2] ^= rstack[step] ^= rstack[step * 2], step = step * 2; 53 else return; 54 } 55 else{ 56 if (rstack[step] >= rstack[step * 2] && rstack[step * 2 + 1] >= rstack[step * 2]) 57 rstack[step] ^= rstack[step * 2] ^= rstack[step] ^= rstack[step * 2], step = step * 2; 58 else if (rstack[step] >= rstack[step * 2 + 1] && rstack[step * 2] >= rstack[step * 2 + 1]) 59 rstack[step] ^= rstack[step * 2 + 1] ^= rstack[step] ^= rstack[step * 2 + 1], step = step * 2 + 1; 60 else return; 61 } 62 } 63 } 64 int main() 65 { 66 int t = 1, n, i; 67 while (scanf("%d", &n) != EOF){ 68 printf("Case %d:\r\n", t++); 69 scanf("%d", &mid); 70 ltot = rtot = 1; 71 printf("%.1lf", (double)mid); 72 for (i = 1; i < n; i++){ 73 if (i % 2){ 74 scanf("%d", &lstack[ltot]); 75 lup(ltot); 76 rdown(); 77 ltot++; 78 printf(" %.1lf", (mid + lstack[1]) / 2.0); 79 } 80 else{ 81 scanf("%d", &rstack[rtot]); 82 rup(rtot); 83 ldown(); 84 rtot++; 85 printf(" %.1lf", (double)mid); 86 } 87 } 88 printf("\r\n"); 89 } 90 return 0; 91 }
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原文地址:http://www.cnblogs.com/zYx-ac/p/4562398.html