1057. Stack (30)

题目链接：http://www.patest.cn/contests/pat-a-practise/1057

题目：

Stack is one of the most fundamental data structures, which is based on the principle of Last In First Out (LIFO). The basic operations include Push (inserting an element onto the top position) and Pop (deleting the top element). Now you are supposed to implement a stack with an extra operation: PeekMedian -- return the median value of all the elements in the stack. With N elements, the median value is defined to be the (N/2)-th smallest element if N is even, or ((N+1)/2)-th if N is odd.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<= 10⁵). Then N lines follow, each contains a command in one of the following 3 formats:

Push key
Pop
PeekMedian

where key is a positive integer no more than 10⁵.

Output Specification:

For each Push command, insert key into the stack and output nothing. For each Pop or PeekMedian command, print in a line the corresponding returned value. If the command is invalid, print "Invalid" instead.

Sample Input:

17PopPeekMedianPush 3PeekMedianPush 2PeekMedianPush 1PeekMedianPopPopPush 5Push 4PeekMedianPopPopPopPop

Sample Output:

InvalidInvalid322124453Invalid

分析：

参考资料：

树状数组http://blog.csdn.net/int64ago/article/details/7429868
http://blog.csdn.net/eli850934234/article/details/8864087
再加二分查找

AC代码：

#include<cstdio>#include<cstring>#include<iostream>#include<string>using namespace std;const int N = 100005;int c[N];int lowbit(int i){//*point_start return i&(-i);}void add(int pos, int value){ while (pos < N){  c[pos] += value;  pos += lowbit(pos); }}int sum(int pos){ int res = 0; while (pos > 0){  res += c[pos];  pos -= lowbit(pos); } return res;}int find(int value){ int l = 0, r = N - 1, median, res; while (l < r - 1){  if ((l + r) % 2 == 0)   median = (l + r) / 2;  else median = (l + r - 1) / 2;  res = sum(median);  if (res < value)   l = median;  else   r = median; } return l + 1;}//*point_endint main(){ freopen("F://Temp/input.txt","r",stdin);   char ss[20]; int stack[N], top = 0, n, pos; memset(c, 0, sizeof(c)); scanf("%d", &n); while (n--){  scanf("%s", ss);  if (ss[1] == 'u'){   scanf("%d", &pos);   stack[++top] = pos;   add(pos, 1);  }  else if (ss[1] == 'o'){   if (top == 0){    printf("Invalid\n");    continue;   }   int out = stack[top];   add(out, -1);   printf("%d\n", stack[top--]);  }  else if (ss[1] == 'e'){   if (top == 0){    printf("Invalid\n");    continue;   }   int res;   if (top % 2 == 0)    res = find(top / 2);   else    res = find((top + 1) / 2);   printf("%d\n", res);  }  else{   printf("Invalid\n");  } } return 0;}

截图：

原来没有用树状数组和二分查找：

后来：

——Apie陈小旭