The 2018 ACM-ICPC Chinese Collegiate Programming Contest Maximum Element In A Stack

 //利用二维数组模拟
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <utility>
 #include <vector>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cstdlib>
 typedef long long ll;
 #define lowbit(x) (x&(-x))
 #define ls l,m,rt<<1
 #define rs m+1,r,rt<<1|1
 using namespace std;
 int t,n,m,p,q;
 const int N=5e6+;
 ll  a[N][];//数据可能超int
 unsigned int SA,SB,SC;
 unsigned int rng61(){
     SA^=SA <<;
     SA^=SA >>;
     SA^=SA <<;
     unsigned int  t=SA;
     SA=SB;
     SB=SC;
     SC ^=t^SA;
     return SC;
 }
 int  main()
 {
     scanf("%d",&t);
     for(int i=;i<=t;i++){
         memset(a,,sizeof(a));
         scanf("%d%d%d%d%u%u%u",&n,&p,&q,&m,&SA,&SB,&SC);
         //每次调用rng61()上面的数会变
         int top=;//因为取最大值时要和a[top-1][1]比较
         //故top初始化为1
         ll ans=;
         for(int i=;i<=n;i++){
             if(rng61()%(p+q)<p){
                 a[top][]=rng61()%m+;//当前应该压入的值
                 a[top][]=max(a[top-][],a[top][]);//令栈顶元素为当前栈内最大值
                 top++;
             }
             else top--;//弹出,而弹出后的栈顶元素仍是弹出后的最大值
             if(top<=) ans^=,top=;//空时不操作
             else {
                 ans^=(i*a[top-][]);//top始终表示栈内元素个数+1
             }
         }
         printf("Case #%d: %lld\n",i,ans);
     }
     return ;
 }