2018ACM-ICPC宁夏邀请赛 A-Maximum Element In A Stack（栈内最大值）

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

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

 
 
 
 
 
int n, p, q, m;
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)
            PUSH(rng61() % m + 1);
        else
            POP();
    }
}
 
 

The procedure PUSH(v) used in the code inserts a new element with value

Output Format

For each test case, output a line containing Case #x: y, where $mathop{oplus}limits_{i = 1}^{n}{left(i cdot a_i ight)}i=1⊕n(i⋅ai) where a_iai is the answer after the ii-th operation and oplus⊕ means bitwise xor.$

Hint

The first test case in the sample input has

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

The second test case also has

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

样例输入

2
4 1 1 4 23333 66666 233333
4 2 1 4 23333 66666 233333

样例输出

Case #1: 19
Case #2: 1

题目来源

The 2018 ACM-ICPC Chinese Collegiate Programming Contest]

思路：向栈中加入当前最大值即可。

8.31update：2019银川网络赛原题。。

#include <bits/stdc++.h>

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<ll> 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
",++tt,ans);
    }
    return 0;
}