1057. Stack (30) (浙大13年机试题)-白红宇

1057. Stack (30) (浙大13年机试题)

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发布时间：2019-06-27

本文共 2954 字，大约阅读时间需要 9 分钟。

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

----------------

开始被这题唬住了。。。。后来仔细想了想树状数组 + 二分乱搞过了1Y。。

View Code

#include 
      
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                  #include 
                   
                    #define CL(arr, val) memset(arr, val, sizeof(arr))#define REP(i, n) for((i) = 0; (i) < (n); ++(i))#define FOR(i, l, h) for((i) = (l); (i) <= (h); ++(i))#define FORD(i, h, l) for((i) = (h); (i) >= (l); --(i))#define L(x) (x) << 1#define R(x) (x) << 1 | 1#define MID(l, r) (l + r) >> 1#define Min(x, y) (x) < (y) ? (x) : (y)#define Max(x, y) (x) < (y) ? (y) : (x)#define E(x) (1 << (x))#define iabs(x) (x) < 0 ? -(x) : (x)#define OUT(x) printf("%I64d\n", x)#define Read() freopen("data.in", "r", stdin)#define Write() freopen("data.out", "w", stdout);typedef long long LL;const double eps = 1e-6;const double PI = acos(-1.0);const int inf = ~0u>>2;using namespace std;const int N = 100010;int c[N];int lowbit(int i) { return i&(-i);}void add(int pos, int val) { while(pos < N) { c[pos] += val; pos += lowbit(pos); }}int sum(int pos) { int res = 0; while(pos > 0) { res += c[pos]; pos -= lowbit(pos); } return res;}char ss[20];int Stack[N], top;int vis[N];int find(int val) { int l = 0, r = N, mid, t; while(r - l > 1) { mid = (l + r) >> 1; t = sum(mid); if(t < val) l = mid; else r = mid; } return l + 1;}int main() { //Read(); CL(c, 0); CL(vis, false); int T, x, top = 0; scanf("%d", &T); while(T--) { scanf("%s", ss); if(strcmp(ss, "Push") == 0) { scanf("%d", &x); add(x, 1); vis[x] ++; Stack[++top] = x; } else { if(top == 0) { puts("Invalid"); continue; } if(ss[1] == 'o') { x = Stack[top--]; add(x, -1); vis[x]--; printf("%d\n", x); } else { printf("%d\n", find((top + 1)/2)); } } } return 0;}

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