题意: t组输入,给你一个长度为n的数组,你每次可以从数组中找到a[i]和a[i+1],然后用a[i]+a[i+1]这个新元素来覆盖掉a[i]和a[i+1]的位置(1<=i<n),从而数组长度也减去1 你可以进行任意次这样的操作,输出最后的数组中有多少数,是p的倍数 题解: 给你一个a数组的前缀和数组w,我们保证a[i]>=1,前缀和w[i]都被p取余过 假设i<j,那么如果w[i]和w[j]相等,那也就是说(a[i+1]+a[i+2]+...+a[j])%p == 0,也就是说这…

作为一个oier,以及大学acm党背包是必不可少的一部分.好久没做背包类动规了.久违地练习下-.- dd__engi的背包九讲:http://love-oriented.com/pack/ 鸣谢http://blog.csdn.net/eagle_or_snail/article/details/50987044,这里有大部分比较有趣的dp练手题. hud 2602 01背包板子题 #include<cstdio> #include<iostream> #include<cs…

Mophues 题意:给出n, m, p,求有多少对a, b满足gcd(a, b)的素因子个数<=p,(其中1<=a<=n, 1<=b<=m) 有Q组数据:(n, m, P <= 5×105. Q <=5000). 参考:ACdreamers 思路:对于hdu1695 GCD来说,由于只需要求gcd = k的个数,所以我们可以按照不优化莫比乌斯公式直接求,这样求解的时间复杂度为O(n); 若这题我们也不优化,答案累加需要双重循环: rep1(i,,n){ if(n…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=4311 Meeting point-1 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3426    Accepted Submission(s): 1131 Problem Description It has been ten years s…

Problem Description Clarke is a patient with multiple personality disorder. One day, Clarke split into two personality a and b, they are playing a game. There is a n∗m matrix, each grid of this matrix has a number ci,j. a wants to beat b every time,…

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 53016    Accepted Submission(s): 20171 Problem Description The least common multiple (LCM) of a set of positive integers is…

题目:Least common multiple 链接:http://acm.hdu.edu.cn/showproblem.php?pid=4913 题意:有一个集合s,包含x1,x2,...,xn,有xi=2^ai * 3^bi,然后给你a数组和b数组,求s所有子集合的最小公倍数之和.比如S={18,12,18},那么有{18},{12},{18},{18,12},{18,18},{12,18},{18,12,18},所以答案是174. 思路: 1. 最小公倍数,因为xi只包含两个质因子2.3…

http://acm.hdu.edu.cn/showproblem.php?pid=6186 题意:给出n个数,共有n次询问,每次询问给出一个数p,求除去第p个数后的n-1个数的&.|.^值. 思路:分别计算出&.|.^的前缀和后缀,将前缀和后缀相计算即可. #include<iostream> #include<cstdio> #include<algorithm> using namespace std; ; int n, q; int a[maxn…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3092 Least common multiple Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 Partychen like to do mathematical problems. One day, when he was doing on a least common m…

题意: 找给定区间的美丽数,美丽数的意思就是这个数每个位上的数都是唯一的. 思路: 前缀和的思想. 感想: 就是你当前位置代表某个特性的前面的所有和(瞎比比的,说了下感觉).前提是你必须找到这样的特性,比如CF的很多题目都是这样子,给你1e5的查询,题解马上一堆线段树,这种区间的预处理,前缀和的思想很好.还有就有一题实现区间的压缩,也是很棒啊. #include <bits/stdc++.h> using namespace std; typedef long long LL; typedef…

sort Time Limit: 6000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25803    Accepted Submission(s): 7764 Problem Description 给你n个整数,请按从大到小的顺序输出其中前m大的数.   Input 每组测试数据有两行,第一行有两个数n,m(0<n,m<1000000),第二行包含n个各不相同…

题目链接: http://codeforces.com/contest/740/problem/D D. Alyona and a tree time limit per test2 secondsmemory limit per test256 megabytes 问题描述 Alyona has a tree with n vertices. The root of the tree is the vertex 1. In each vertex Alyona wrote an positiv…

Lowest Common Multiple Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34980    Accepted Submission(s): 14272 Problem Description 求n个数的最小公倍数.   Input 输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数.  …

Problem Description The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.   Input Input will consist of multiple pr…

题目链接 \(Description\) 给定一张\(n\)个点\(m\)条边的无向图.定义割集\(E\)为去掉\(E\)后使得图不连通的边集.定义一个bond为一个极小割集(即bond中边的任意一个真子集都不是割集). 对每条边,求它在多少个bond中. \(n\leq20,\quad n-1\leq m\leq\frac{n(n-1)}{2}\). \(Solution\) https://www.cnblogs.com/zufezzt/p/5723389.html 首先bond是极小割集,…

http://acm.hdu.edu.cn/showproblem.php?pid=2028 Problem Description 求n个数的最小公倍数.   Input 输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数.   Output 为每组测试数据输出它们的最小公倍数,每个测试实例的输出占一行.你可以假设最后的输出是一个32位的整数.   Sample Input 2 4 6 3 2 5 7   Sample Output 12 70   代码: #includ…

Twin Prime Conjecture Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description If we define dn as: dn = pn+1-pn, where pi is the i-th prime. It is easy to see that d1 = 1 and dn=even for n>1. Twin Prime…

题意: 给你一串字符串,每个字符都有一个权值,要求把这个字符串在某点分开,使之成为两个单独的字符串 如果这两个子串某一个是回文串,则权值为那一个串所有的字符权值和 若不是回文串,则权值为0 解析: 先用Manacher算法求出以每个字母为中心的回文串的长度,并计算该字符串的前缀价值和.然后枚举切割点,得到两份子串.这样就可以知道每个子串的中心点,然后检查以该子串的中心点作为中心点的回文串的长度,如果长度等于该子串的长度,那么就加上该子串的价值.然后和最优价值比较就行了. 注意只有一个字符时输出0…

进制转换 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 63039    Accepted Submission(s): 34261 Problem Description 输入一个十进制数N,将它转换成R进制数输出.   Input 输入数据包含多个测试实例,每个测试实例包含两个整数N(32位整数)和R(2<=R<=16, R<…

Olympiad Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description You are one of the competitors of the Olympiad in numbers. The problem of this year relates to beatiful numbers. One integer is called be…

Phone List Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16223    Accepted Submission(s): 5456 Problem Description Given a list of phone numbers, determine if it is consistent in the sense tha…

Game Time Limit:1500MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 5242 Appoint description:  System Crawler  (2015-05-26) Description It is well known that Keima Katsuragi is The Capturing God because of his…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4055 题意: 给你一个由'I', 'D', '?'组成的字符串,长度为n,代表了一个1~n+1的排列中(数字不重复),相邻数字的增长趋势.('I'为增,'D'为减,'?'为未知) 问你符合条件的数列有多少种. 题解: 表示状态: dp[i][j] = combinations 表示长度为i的排列(由1~i组成),末尾为j,这样的排列的个数 找出答案: ans = ∑ dp[n][1 to n] 如何…

Description Partychen like to do mathematical problems. One day, when he was doing on a least common multiple(LCM) problem, he suddenly thought of a very interesting question: if given a number of S, and we divided S into some numbers , then what is…

题目描述 区间质数个数 输入输出格式 输入格式: 一行两个整数 询问次数n,范围m 接下来n行,每行两个整数 l,r 表示区间 输出格式: 对于每次询问输出个数 t,如l或r∉[1,m]输出 Crossing the line 输入输出样例 输入样例#1: 复制 2 5 1 3 2 6 输出样例#1: 复制 2 Crossing the line 说明 [数据范围和约定] 对于20%的数据 1<=n<=10 1<=m<=10 对于100%的数据 1<=n<=1000 1…

http://acm.hdu.edu.cn/showproblem.php?pid=1019 LCM即各数各质因数的最大值,搞个map乱弄一下就可以了. #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned int ui; map<ui,ui> M; ll _pow(ui f,ui s){ ll res=1; while(s){ res*=f; s--; } retur…

Yet Another Multiple Problem Time Limit: 40000/20000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3407    Accepted Submission(s): 825 Problem Description There are tons of problems about integer multiples. Despit…

http://acm.hdu.edu.cn/showproblem.php?pid=2227 用dp[i]表示以第i个数为结尾的nondecreasing串有多少个. 那么对于每个a[i] 要去找 <= a[i]的数字那些位置,加上他们的dp值即可. 可以用树状数组维护 #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algori…

点击加号查看代码 #include<bits/stdc++.h>//前缀和优化版本,不易理解 using namespace std; #define ll long long ; ; ll sum[maxn][maxn]; ll dp[maxn][maxn]; char str[maxn]; int main() { str[]='*'; str[]='*'; scanf(); ll len=strlen(str)-; sum[][]=; dp[][]=; ;i<=len;i++) ;…

B. Balanced Substring You are given a string s consisting only of characters 0 and 1. A substring [l, r] of s is a string slsl + 1sl + 2... sr, and its length equals to r - l + 1. A substring is called balanced if the number of zeroes (0) equals to t…