How many integers can you find Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1796 Description   Now you get a number N, and a M-integers set, you should find out how many integers which are sm…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4249    Accepted Submission(s): 1211 Problem Description   Now you get a number N, and a M-integers set, you shoul…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3315    Accepted Submission(s): 937 Problem Description   Now you get a number N, and a M-integers set, you shoul…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5664    Accepted Submission(s): 1630 Problem Description   Now you get a number N, and a M-integers set, you shoul…

题意:找出小于n是m个数每个数的倍数的数的个数. 思路:用二进制表示是那几个数的倍数. 二进制进行容斥,去掉小于0的数. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; __int64 n,m,g; __int64 a[],b[]; __int64 gcd(__int64 a,__int64 b) { ?a:gcd(b,a%b); } int main() {…

题目传送:http://acm.hdu.edu.cn/diy/contest_showproblem.php?cid=20918&pid=1002 Problem Description   Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers…

HDU.1796 How many integers can you find ( 组合数学 容斥原理 二进制枚举) 题意分析 求在[1,n-1]中,m个整数的倍数共有多少个 与 UVA.10325 The Lottery 一模一样. 前置技能和其一样,但是需要注意的有一下几点: 1. m个数字中可能有0 2. 要用long long 代码总览 #include <cstdio> #include <algorithm> #include <cstring> #incl…

How many integers can you find Problem Description   Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-intege…

题目链接 题意 : 给你N,然后再给M个数,让你找小于N的并且能够整除M里的任意一个数的数有多少,0不算. 思路 :用了容斥原理 : ans = sum{ 整除一个的数 } - sum{ 整除两个的数 } + sum{ 整除三个的数 }………………所以是奇加偶减,而整除 k 个数的数可以表示成 lcm(A1,A2,…,Ak) 的倍数的形式.所以算出最小公倍数, //HDU 1796 #include <cstdio> #include <iostream> #include <…

A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4032    Accepted Submission(s): 1255 Problem Description Let A1, A2, ... , AN be N elements. You need to deal with…

A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5339    Accepted Submission(s): 1693 Problem Description Let A1, A2, ... , AN be N elements. You need to deal with…

题目连接   http://acm.hdu.edu.cn/showproblem.php?pid=1796 处男容斥原理  纪念一下  TMD看了好久才明白DFS... 先贴代码后解释 #include<cstdio> #include<cstring> using namespace std; #define LL long long #define N 11 LL num[N],ans,n; int m,cnt; LL gcd(LL a,LL b) { int t; while…

A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5402    Accepted Submission(s): 1710 Problem Description Let A1, A2, ... , AN be N elements. You need to deal with…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description   Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that…

A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4000    Accepted Submission(s): 1243 Problem Description Let A1, A2, ... , AN be N elements. You need to deal with…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6434    Accepted Submission(s): 1849 Problem Description   Now you get a number N, and a M-integers set, you shou…

http://acm.hdu.edu.cn/showproblem.php?pid=4267 [思路] 树状数组的区间修改:在区间[a, b]内更新+x就在a的位置+x. 然后在b+1的位置-x 树状数组的单点查询:求某点a的值就是求数组中1~a的和. (i-a)%k==0把区间分隔开了,不能直接套用树状数组的区间修改单点查询 这道题的K很小,所以可以枚举k,对于每个k,建立k个树状数组,所以一共建立55棵树 所以就可以多建几棵树..然后就可以转换为成段更新了~~ [AC] #include<b…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5556    Accepted Submission(s): 1593 Problem Description   Now you get a number N, and a M-integers set, you shou…

How many integers can you find Time Limit: 12000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6710    Accepted Submission(s): 1946 Problem Description   Now you get a number N, and a M-integers set, you shou…

A Simple Problem with Integers Case Time Limit: 2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other i…

the algorithm of three version below is essentially the same, namely, Kadane's algorithm, which is of O(n) complexity. https://en.wikipedia.org/wiki/Maximum_subarray_problem the evolution of the implementations is to remove redundancy and do what is…

题意:给一个序列,操作1:给区间[a,b]中(i-a)%k==0的位置 i 的值都加上val  操作2:查询 i 位置的值 解法:树状数组记录更新值. 由 (i-a)%k == 0 得知 i%k == a%k,又因为k <= 10,想到建55棵树状数组,即对每个(k,x%k)都建一棵树状数组,每次更新时,在第(k,a%k)棵树状数组上更新a这个点,更新值为val,然后再b+1处更新值为-val,即在[a,b]内更新了val. 查询pos的时候,求出每一个树状数组(k,pos%k)的sum值即可.…

容斥原理!! 这题首先要去掉=0和>=n的值,然后再使用容斥原理解决 我用的是数组做的…… #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<string> #include<vector> #define ll __int64 using namespace std;…

题意: 2种操作 1 a b k c 在区间[a,b]中的(i-a)%k==0的位置i上的数+c 2 a 查询位置a的值 输出每次查询的值 分析: 开始想到多维的线段树,但比较麻烦,看了题解才知道,用BIT实现区间更新,单点查询,若在区间[a,b]上的数加c 就在a位置加c ,b+1位置加-c 这样在查询时sum(i),(i>=a&&i<=b)就是当前i位置的值 维护多个BIT即可 #include <map> #include <set> #inclu…

<题目链接> 题目大意: 给你m个数,其中可能含有0,问有多少小于n的正数能整除这个m个数中的某一个. 解题分析: 容斥水题,直接对这m个数(除0以外)及其组合的倍数在[1,n)中的个数即可,因为可能会重复计算,所以在叠加的时候进行容斥处理,下面用的是位运算实现容斥. #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namesp…

思路:二进制解决容斥问题,就和昨天做的差不多.但是这里题目给的因子不是质因子,所以我们求多个因子相乘时要算最小公倍数.题目所给的因数为非负数,故可能有0,如果因子为0就要删除. 代码: #include<set> #include<map> #include<cmath> #include<queue> #include<cstdio> #include<cstring> #include<iostream> #inclu…

题意:给定一个数 n,和一个集合 m,问你小于的 n的所有正数能整除 m的任意一个的数目. 析:简单容斥,就是 1 个数的倍数 - 2个数的最小公倍数 + 3个数的最小公倍数 + ...(-1)^(n+1) * n个数的最小公倍数. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdl…

以前似乎做过类似的不过当时完全不会.现在看到就有点思路了,开始还有洋洋得意得觉得自己有不小的进步了,结果思路错了...改了很久后测试数据过了还果断爆空间... 给你一串数字A,然后是两种操作: "1 l r k c":意思是当 l=<i<=r 对(i-a)%k = =0 的每个 Ai 都增加 c (1=<k<=10) "2 i" :意思是求出 Ai 一看就是区间更新和单点查询,其实可以用树状数组做,可是觉得线段树好弄一点,结果成功入坑...…

题意: 给一个N.然后给M个数,问1~N-1里面有多少个数能被这M个数中一个或多个数整除. 思路: 首先要N-- 然后对于每一个数M 事实上1~N-1内能被其整除的 就是有(N-1)/M[i]个 可是会出现反复 比方 例子 6就会被反复算 这时候我们就须要容斥原理了 加上一个数的减去两个数的.. 这里要注意了 两个数以上的时候 是求LCM而不是简单的相乘! 代码: #include "stdio.h" #include "string.h" #include &qu…

Code: #include<cstdio> using namespace std; typedef long long ll; const int R=13; ll a[R]; ll n,ans; int m,cnt=0; ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);} void dfs(int cur,ll lcm,int id){ if(cur>cnt)return; lcm=a[cur]/gcd(a[cur],lcm)*lcm; if(…