hdu-5587 Array(递归)】的更多相关文章

题目链接:hdu 5587 前两周 bc 上的题了,因为赶大作业所以没有去打,看了下官方给出的思路,感觉好强大~~竟然能转化成求二进制数 1 的个数: 然后数位 dp 就行了, #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef unsigned long long ull; #define For(i,s,t) for(int i = s; i <…
Array Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5587 Description Vicky is a magician who loves math. She has great power in copying and creating.One day she gets an array {1}. After that, every day she cop…
Array Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Vicky is a magician who loves math. She has great power in copying and creating.One day she gets an array {1}. After that, every day she c…
Array Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 417    Accepted Submission(s): 211 Problem Description Vicky is a magician who loves math. She has great power in copying and creating.One…
题目大意:给出一个n个元素的数组A,A中所有元素都是不重复的[1,n].有两种操作:1.将pos位置的元素+1e72.查询不属于[1,r]中的最小的>=k的值.强制在线. 题解因为数组中的值唯一,且在1到n的范围内,而询问的r和k也在1到n的范围内. 所以对于任意一个被操 作1修改过的值都不会成为询问的答案,而询问的结果也必然在k到n+1的范围内. 因为没有被修改过 值是唯一的,所以可以建立权值线段树,维护权值区间内的值所在下标的最大值.而询问则转化为不小 于k的值里面,下标超过r的最小权值是多…
Array  Accepts: 118  Submissions: 232  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others) 问题描述 Vicky是个热爱数学的魔法师,拥有复制创造的能力. 一开始他拥有一个数列{1}.每过一天,他将他当天的数列复制一遍,放在数列尾,并在两个数列间用0隔开.Vicky想做些改变,于是他将当天新产生的所有数字(包括0)全加1.Vicky现在想考考你…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6197 题意:给你n个数,问让你从中删掉k个数后(k<=n),是否能使剩下的序列为非递减或者非递增序列 解法:签到题,就是让你求最长不下降子序列长度len,然后判断下n-len是否小于k(将序列反着存下来然后再求即最长不上升子序列,取两者len中的较大值),然后直接套nlogn的模板即可. #include <bits/stdc++.h> using namespace std; const…
因为n>1000000000所以用递归 数组超限, 由递归函数f(n)=(A*f(n-1)+B*f(n-2))%7; 因为是除7的余数 因次一共有7*7=49种情况, 以后的值都和之前的对应相等 所以找到和3,4 重复的数之后,就很容易得到f(n)…
array array array Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 459    Accepted Submission(s): 282 Problem Description One day, Kaitou Kiddo had stolen a priceless diamond ring. But detective…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6172 题意:如题. 解法: #include <bits/stdc++.h> using namespace std; typedef long long LL; const LL mod = 1e9+7; struct Matrix{ LL a[3][3]; void set1(){ memset(a, 0, sizeof(a)); } void set2(){ memset(a, 0, siz…