题目链接:hdu_5286_wyh2000 and sequence 题意: 给一段长度为N的序列,每次询问l-r(l和r和上一次询问的答案有关)内 不同的数的 出现次数的次方 的和.强制在线 题解: 这里贴个达哥的题解: 大体思路就是,把n个数分成sqrt(n)块,每块sqrt(n)个数,然后求出任意两块i到j(包含第i块和第j块)的ans,对于每次询问l和r,找到刚好包含l和r的一段连续的块,因为我们已经知道了这段连续块的答案,所以只要再去掉两段多出来的一部分,就好了. 为了得到任意两块间的…

Argestes and Sequence Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 566    Accepted Submission(s): 142 Problem Description Argestes has a lot of hobbies and likes solving query problems espec…

Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Argestes has a lot of hobbies and likes solving query problems especially. One day Argestes came up with such a problem. Y…

给定递推式: 求Fn. 分析:给出的公式可以用快速矩阵幂运算得到,但 P/n 整除对于不同的i,值是不同的. 可以根据P将3-n分成若干块,每块中P整除n的值是相同的.分块的时候要注意判断. 将每块的快速幂结果累乘得到结果. #include <bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<int, int> PII; ; ; ; int N; int A, B, C, D, P; in…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6395 因为题目数据范围太大,又存在递推关系,用矩阵快速幂来加快递推. 每一项递推时  加的下取整的数随着n变化,但因为下取整有连续性(n一段区间下取整的数是相同的),可以分块,相同的用矩阵快速幂加速 想了好久..如果最小的开始的值是[p/i]的数为i,那连续的一段长度是[p/(p/i)]-i+1,但为什么分段数是根号n级别啊?... 套矩阵快速幂,时间复杂度O(sqrt(n) * log(n)) ⎧…

题目链接 #include<bits/stdc++.h> using namespace std; #define e exp(1) #define pi acos(-1) #define mod 1000000007 #define inf 0x3f3f3f3f #define ll long long #define ull unsigned long long #define mem(a,b) memset(a,b,sizeof(a)) int gcd(int a,int b){retu…

Sequence Problem Description Let us define a sequence as below f1=A f2=B fn=C*fn-2+D*fn-1+[p/n] Your job is simple, for each task, you should output Fn module 109+7.   Input The first line has only one integer T, indicates the number of tasks. Then,…

Sequence Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1731    Accepted Submission(s): 656 Problem Description Let us define a sequence as below F1=A F2=B Fn=C⋅Fn−2+D⋅Fn−1+⌊Pn⌋ Your job is s…

HDU5286 wyh2000 and sequence 题意: 给出长为\(N\)的序列\(A_1,A_2,A_3,\cdots,A_n\),\(q\)次询问,每次询问给出区间\([L,R]\),假设区间内出现过的数为\(C_1,C_2,\cdots,C_k\),出现的次数分别为\(B_1,B_2,\cdots,B_k\),输出\(\sum_{i=1}^{k}C_i^{B_i} % 1000000007\),要求在线 题解: 如果不要求在线,直接用莫队很快就能解决,现在要求在线,还是考虑把序列…

Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35049   Accepted: 10139   Special Judge Description Let us define a regular brackets sequence in the following way: 1. Empty sequence is a regular sequence. 2. If S is a r…