在使用单调栈的过程中,栈中保留的奶牛 都是比当前 值 大的奶牛,这时记录此时栈中的奶牛数量,就是在当前奶牛的前面所有奶牛当中可以看到当前奶牛的奶牛数量。。。。好绕口。。。。。 #include<cstdio>#include<cstring>using namespace std;long long int stack[88888];long long d[88888];int m
Bad Hair Day Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 48 Accepted Submission(s) : 17 Problem Description Some of Farmer Jo
题目: Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see th
单调栈版子 1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 const int N=80010; 5 int n,a,top=0,zh[N]; 6 int read(){ 7 int sum=0; 8 char ch=getchar(); 9 while (ch<'0
文章目录 题168.洛谷P2866 单调栈-Bad Hair Day S一、题目二、题解 题168.洛谷P2866 单调栈-Bad Hair Day S 一、题目 二、题解 题目要你计算牛往右看,每头牛能看到牛顶的数目之和。依题意我们想,往又右看,也就是只要右边的牛都比当前牛严格矮,那么就能看到牛顶,如果碰到一个不满足这个条件的牛直接让当前牛话下一个,不用再
题目: Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the to