Maximum Element In A Stack

•  20.91%
•  10000ms
•  262144K

 

As an ACM-ICPC newbie, Aishah is learning data structures in computer science. She has already known that a stack, as a data structure, can serve as a collection of elements with two operations:

• push, which inserts an element to the collection, and
• pop, which deletes the most recently inserted element that has not yet deleted.

Now, Aishah hopes a more intelligent stack which can display the maximum element in the stack dynamically. Please write a program to help her accomplish this goal and go through a test with several operations.

Aishah assumes that the stack is empty at first. Your program will output the maximum element in the stack after each operation. If at some point the stack is empty, the output should be zero.

Input Format

The input contains several test cases, and the first line is a positive integer TT indicating the number of test cases which is up to 5050.

To avoid unconcerned time consuming in reading data, each test case is described by seven integers n~(1\le n\le 5 \times 10^6)n (1≤n≤5×106), pp, qq, m~(1\le p, q, m\le 10^9)m (1≤p,q,m≤109), SASA, SBSB and SC~(10^4 \le SA, SB, SC\le 10^6)SC (104≤SA,SB,SC≤106).The integer nn is the number of operations, and your program should generate all operations using the following code in C++.

 

 

 

 

 

1

int n, p, q, m;

2

unsigned int SA, SB, SC;

3

unsigned int rng61(){

4

   SA ^= SA << 16;

5

   SA ^= SA >> 5;

6

   SA ^= SA << 1;

7

   unsigned int t = SA;

8

   SA = SB;

9

   SB = SC;

10

   SC ^= t ^ SA;

11

   return SC;

12

}

13

void gen(){

14

   scanf("%d%d%d%d%u%u%u", &n, &p, &q, &m, &SA, &SB, &SC);

15

   for(int i = 1; i <= n; i++){

16

       if(rng61() % (p + q) < p)

17

           PUSH(rng61() % m + 1);

18

       else

19

           POP();

20

   }

21

}

 

 

The procedure PUSH(v) used in the code inserts a new element with value vv into the stack and the procedure POP() pops the topmost element in the stack or does nothing if the stack is empty.

Output Format

For each test case, output a line containing Case #x: y, where xx is the test case number starting from 11, and yy is equal to \mathop{\oplus}\limits_{i = 1}^{n}{\left(i \cdot a_i\right)}i=1⊕n​(i⋅ai​) where a_iai​ is the answer after the ii-th operation and \oplus⊕ means bitwise xor.

Hint

The first test case in the sample input has 44 operations:

• POP();
• POP();
• PUSH(1);
• PUSH(4).

The second test case also has 44 operations:

• PUSH(2);
• POP();
• PUSH(1);
• POP().

样例输入

24 1 1 4 23333 66666 2333334 2 1 4 23333 66666 233333

样例输出

Case #1: 19Case #2: 1

题目来源

]

 

 

向栈中加入当前最大值即可。

 

#include 
     
      using namespace std;typedef long long ll;const int MAX = 100005;const int INF = 0x3f3f3f3f;const int MOD = 1000000007;int n, p, q, m;ll ans;stack
      
        s;unsigned int SA, SB, SC;unsigned int rng61(){    SA ^= SA << 16;    SA ^= SA >> 5;    SA ^= SA << 1;    unsigned int t = SA;    SA = SB;    SB = SC;    SC ^= t ^ SA;    return SC;}void gen(){    scanf("%d%d%d%d%u%u%u", &n, &p, &q, &m, &SA, &SB, &SC);    for(int i = 1; i <= n; i++){        if(rng61() % (p + q) < p){            int x=rng61() % m + 1;            if(!s.size()) s.push(x);            else if(x>s.top()) s.push(x);            else s.push(s.top());        }        else{            if(s.size()) s.pop();            else continue;        }        if(s.size()) ans^=i*s.top();    }}int main(void){    int t,tt=0,i;    scanf("%d",&t);    while(t--){        while(s.size()) s.pop();        ans=0;        gen();        printf("Case #%d: %lld\n",++tt,ans);    }    return 0;}