https://nanti.jisuanke.com/t/41285
题干：
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
The input contains several test cases, and the first line is a positive integer TTindicating 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≤n≤5×106)n (1≤n≤5×106), pp, qq, m (1≤p,q,m≤109)m (1≤p,q,m≤109), SASA, SBSB and SC (104≤SA,SB,SC≤106)SC (104≤SA,SB,SC≤106). The integer nn is the number of operations, and your program is asked to generate all operations by 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(" for(int i = 1; i <= n; i++){ if(rng61() PUSH(rng61() else POP(); } }
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
For each test case, output a line containing Case #x: y, where x is the test case number starting from 11, and y is equal to ⊕i=1n(i⋅ai)⊕i=1n(i⋅ai) where ⊕⊕ means bitwise xor.
Example
Input
2
4 1 1 4 23333 66666 233333
4 2 1 4 23333 66666 233333
Output
Case #1: 19
Case #2: 1
Note
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().
题目大意：
刚开始是一个空栈， 每次可以在其中加入元素，并且要求出此时栈中元素的最大值。对于每个样例只需要输出一个数：
解题报告：
对于push操作，读入一个数x，每次加入sk.top()和x的较大值，这样保证栈顶元素一定是最大值。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
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;
}ll gen() {stack<ll> sk;ll res = 0;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) {ll tmp = rng61() %m + 1;if(sk.empty()) sk.push(tmp);else sk.push(max(tmp,sk.top()));}else if(sk.size()) sk.pop();if(sk.size()) res ^= i*sk.top();else res ^= 0;}return res;
}
int main()
{int t,iCase=0;cin>>t;while(t--) {printf("Case #%d: %lld\n",++iCase,gen());}return 0 ;
}
