2015 HDU 多校联赛 5317 RGCDQ 筛法求解 题目  http://acm.hdu.edu.cn/showproblem.php? pid=5317 本题的数据量非常大,測试样例多.数据量大, 所以必须做预处理.也就是用筛法求出全部的F[x],将全部F[x] 打印出来发现.事实上结果不大,最大的数值是7.所以对于每一个区间询问, 直接暴力求取有多少个 1 2 3 4 5 6 7 就可以,从大到小查找.假设出现2个以上 3-7 的数值,那么最大公约数就是该数字. 假设没有出现两个反复…

2015 HDU 多校联赛 5363 Key Set 题目: http://acm.hdu.edu.cn/showproblem.php? pid=5363 依据前面给出的样例,得出求解公式 fn = 2^(n-1) - 1, 数据量大,实际就是求幂次方. 可用分治法求解.复杂度O(nlogn) // 分治法求高速幂 #include <bits/stdc++.h> using namespace std; typedef unsigned long long ull; const int M…

题目传送门 /* 题意:给一个区间,问任意两个数的素数因子的GCD最大 数学+dp:预处理出f[i],发现f[i] <= 7,那么用dp[i][j] 记录前i个f[]个数为j的数有几个, dp[r][j] - dp[l-1][j]表示区间内j的个数,情况不多,分类讨论一下 */ #include <cstdio> #include <algorithm> #include <cstring> #include <vector> #include <…

2018 HDU多校第四场赛后补题 自己学校出的毒瘤场..吃枣药丸 hdu中的题号是6332 - 6343. K. Expression in Memories 题意: 判断一个简化版的算术表达式是否合法. 题解: 注意细节即可. 代码: #include <bits/stdc++.h> using namespace std; int n; char s[505]; int main () { int T; cin>>T; for ( ; T; --T) { scanf(&quo…

2018 HDU多校第三场赛后补题 从易到难来写吧,其中题意有些直接摘了Claris的,数据范围是就不标了. 如果需要可以去hdu题库里找.题号是6319 - 6331. L. Visual Cube 题意: 在画布上画一个三维立方体. 题解: 模拟即可. 代码: #include <bits/stdc++.h> using namespace std; int a, b, c, R, C; char g[505][505]; int main () { int T; cin >>…

[HDU多校]Ridiculous Netizens 点分治 分成两个部分:对某一点P,连通块经过P或不经过P． 经过P采用树形依赖背包 不经过P的部分递归计算 树型依赖背包 v点必须由其父亲u点转移过来 即必须经过P点 \(dp[v][s*a[v]]+=dp[u][s]\) 第二维代表连通块的乘积 第一维代表经过该点并且一定经过P点的方案数 所以最后父节点还要加上子节点的方案数 空间优化 第二维不能开这么大 稍稍转变含义,改成还能＂装下＂多少体积 \(\lfloor \frac{m}{s}\r…

题目链接:pid=5317" target="_blank">http://acm.hdu.edu.cn/showproblem.php?pid=5317 Problem Description Mr. Hdu is interested in Greatest Common Divisor (GCD). He wants to find more and more interesting things about GCD. Today He comes up with…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5317 题意:F(x) 表示x的不同质因子的个数结果是求L,R区间中最大的gcd( F(i) , F(j) ),i.j在L,R区间内. 思路:因为2<=L < R<=1000000,所以他们的质因子最多的个数也就7个,也就是说1<=F(x)<=7,因为要求最大的GCD,所以只要知道在L,R区间内每个F(x)的情况就可以知道结果. 代码: #include <stdio.h&g…

题目连接:Click here 题意:在一个[L,R]内找到最大的gcd(f[i],f[j])其中L<=i<j<=R,f[x]表示i分解质因数后因子的种类数.eg:f[10]=2(10=2*5),f[12]=2(12=2*2*3). 分析:很容易想到先将f[x]求出来,这里x最大1e6,要在常数时间内求出f[x].并且稍加分析就知道1<=f[x]<=7,可以用一个dp[i][j]表示从f[1]到f[i]有多少个j.这样就可以在常数时间内预处理出来,后面在O(1)的时间内就可以…

RGCDQ Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1652    Accepted Submission(s): 696 Problem Description Mr. Hdu is interested in Greatest Common Divisor (GCD). He wants to find more and mo…

题意:从1~1000,000的每个自然数质因子分解,不同因子的个数作为其f 值,比如12=2*2*3,则f(12)=2.将100万个数转成他们的f值后变成新的序列seq.接下来T个例子,每个例子一个询问区间seq[L,R].问该子序列中任意两个不同下标的数,他们的GCD值最大为多少? 思路: (1)质因子分解,用筛法先将质数先全部出来. (2)对每个自然数求其f值,即将其质因子分解后的不同因子个数,作为seq序列. (3)扫一遍seq,用另一个数组记录下标i之前有多少个1.2.3.4....因为…

RGCDQ Time Limit: 3000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status Description Mr. Hdu is interested in Greatest Common Divisor (GCD). He wants to find more and more interesting things about GCD. Today He comes up with Ra…

题目链接:hdu 5317 这题看数据量就知道需要先预处理,然后对每个询问都需要在 O(logn) 以下的复杂度求出,由数学规律可以推出 1 <= F(x) <= 7,所以对每组(L, R),只需要求出它们在 1~7 的范围内的数量,然后暴力求出 gcd 即可.因为符合递增,可以设一个结点 struct { v[8]; } 记录 1~7 的数量,结点间可以相加减,也就可以用前缀和的思想去做(其实我也是看了别人的题解才明白这种思想,一开始用二分不是超时就是 wa 了,不过我竟发现自己手写的二分比…

Painter Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 895    Accepted Submission(s): 408 Problem Description Mr. Hdu is an painter, as we all know, painters need ideas to innovate , one day,…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 593    Accepted Submission(s): 164 Problem Description There are n circles on a infinitely large table.With every two circle, either one contains…

Problem Description Mr. Hdu is interested in Greatest Common Divisor (GCD). He wants to find more and more interesting things about GCD. Today He comes up with Range Greatest Common Divisor Query (RGCDQ). What’s RGCDQ? Please let me explain it to you…

题意:f(i)表示i的质因子个数,给l和r,问在这一区间内f(i)之间任意两个数最大的最大公倍数是多少. 解法:先用筛法筛素数,在这个过程中计算f(i),因为f(i)不会超过7,所以用一个二维数组统计前i个数中每个f(i)出现的次数,当询问l和r时,用num[r] - num[l - 1],得到这一区间内的结果,然后讨论一下,如果出现过6和3则答案可能为3,如果出现过4和2则答案可能为2,如果某数字出现两次及以上则答案可能是这个数字,以上几种情况取最大值,即为答案. 代码: #include<s…

Virtual Participation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 237    Accepted Submission(s): 56 Special Judge Problem Description As we know, Rikka is poor at math. Yuta is worrying abo…

Connect the Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 108    Accepted Submission(s): 36 Special Judge Problem Description Once there was a special graph. This graph had n vertices a…

Tricks Device Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1977    Accepted Submission(s): 509 Problem Description Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu’s at the en…

Crazy Bobo Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1215    Accepted Submission(s): 366 Problem Description Bobo has a tree,whose vertices are conveniently labeled by 1,2,...,n.Each nod…

Fantasy magicians usually gain their ability through one of three usual methods: possessing it as an innate talent, gaining it through study and practice, or receiving it from another being, often a god, spirit, or demon of some sort. Some wizards ar…

Solve this interesting problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 422    Accepted Submission(s): 98 Problem Description Have you learned something about segment tree? If not, don’t…

Your current task is to make a ground plan for a residential building located in HZXJHS. So you must determine a way to split the floor building with walls to make apartments in the shape of a rectangle. Each built wall must be paralled to the buildi…

Delicious Apples Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 1371    Accepted Submission(s): 448 Problem Description There are n apple trees planted along a cyclic road, which is L metres…

Friends Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 163    Accepted Submission(s): 61 Problem Description There are n people and m pairs of friends. For every pair of friends, they can choo…

OO’s Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 449    Accepted Submission(s): 158 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the nu…

1.题目描写叙述:pid=5411">点击打开链接 2.解题思路:本题实际是是已知一张无向图.问长度小于等于m的路径一共同拥有多少条. 能够通过建立转移矩阵利用矩阵高速幂解决.当中,转移矩阵就是输入时候的邻接矩阵,同一时候多添加最后一列,都置为1.表示从i開始的,长度不超过M的路径的答案总数(最后一行的1~n列为全0行,能够理解为空集),那么把转移矩阵自乘M-1次后就是路径长度为M的转移矩阵(这里的路径长度指的是顶点的个数.顶点=边数+1,因此仅仅须要乘M-1次). 为何便于求和.能够设置…

CRB has a tree, whose vertices are labeled by 1, 2, …, N. They are connected by N – 1 edges. Each edge has a weight. For any two vertices u and v(possibly equal), f(u,v) is xor(exclusive-or) sum of weights of all edges on the path from u to v. CRB’s…

Problem Description CRB has two strings s and t. In each step, CRB can select arbitrary character c of s and insert any character d (d ≠ c) just after it. CRB wants to convert s to t. But is it possible?   Input There are multiple test cases. The fir…