传送门:点击打开链接 题意:有n个箱子排成一排,有m个炸弹.位置告诉你.如今炸弹的左边伤害和右边伤害能够自己控制,要求 每一个炸弹炸的箱子数的累乘,输出答案取log2并乘以1e6 思路:直接2for xjb搞即可了.大概就是某个区间里刚好仅仅有一个炸弹时,就是满足的,然后就从前面往后面更新一下 #include <map> #include <set> #include <cmath> #include <ctime> #include <stack&…

xiaoxin and his watermelon candy 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5654 Description During his six grade summer vacation, xiaoxin got lots of watermelon candies from his leader when he did his internship at Tencent. Each watermelon cand…

pid=5654">[HDOJ 5654] xiaoxin and his watermelon candy(离线+树状数组) xiaoxin and his watermelon candy Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 233    Accepted Submission(s): 61 Problem Des…

xiaoxin and his watermelon candy Problem Description During his six grade summer vacation, xiaoxin got lots of watermelon candies from his leader when he did his internship at Tencent. Each watermelon candy has it's sweetness which denoted by an inte…

Problem Description During his six grade summer vacation, xiaoxin got lots of watermelon candies from his leader when he did his internship at Tencent. Each watermelon candy has it's sweetness which denoted by an integer number. xiaoxin is very smart…

题目链接 题意:序列长度为n(1<= n <= 200,000)的序列,有Q(<=200,000)次区间查询,问区间[l,r]中有多少个不同的连续递增的三元组. 思路:连续三元组->递推O(n)将第一次出现该三元组的下标记录到树状数组中,并且用一个Next[]来表示递推关系,即同一个三元组下一次出现的位置是Next[x];这很关键,在之后按照左边界处理时(有点像莫队算法),一直递推在l+1左侧重复出现的三元组,为了把该三元组推到出现在[l,r]或者[r+1,..]中,这样不重不漏.…

题目链接 求给出的区间中有多少个三元组满足i+1=j=k-1 && a[i]<=a[j]<=a[k] 如果两个三元组的a[i], a[j], a[k]都相等, 那么这两个三元组算一个. 预处理一下所有三元组, 然后跑莫队就水过去了... #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm&…

传送门:点击打开链接 题意:一个三元组假设满足j=i+1,k=j+1,ai<=aj<=ak,那么就好的.如今告诉你序列.然后Q次询问.每次询问一个区间[l,r],问区间里有多少个三元组满足要求 思路:刚開始看错题目了,原来三元组是连续3个,这作为bc最后一题也太水了把. . . 先一遍预处理.把连续3个满足条件的找出来,放到还有一个数组里排序去重,用这个数组来给三元组哈希.再扫一遍给三元组在之前那个排序好的数组里二分一下得到下标,大概就是哈希一下,用一个数字来表示. 之后的查询.事实上就是.在…

题意:bc 77div1 d题(中文题面),其实就是询问一个区间有多少不同的三元组,当然这个三元组要符合条件 分析(先奉上官方题解) 首先将数列中所有满足条件的三元组处理出来,数量不会超过 nn个. 设 pre[i] 为第 i 个三元组前一次出现的位置,如果在前面没有出现过则设为0,对于不合法的三元组,pre[i]设为 n. 这样对于一个查询 [L, R], 我们只不需要计算出 在这个区间内有多少个三元组的 pre 值是小于 L 的. 到这里,就可以使用可持久化线段树来计算了. --------…

T1 xiaoxin juju needs help 计算组合数然后多重集排列乱搞,注意判无解情况(TM我就判错然后FST了). #include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) using…