题意: 给出n, k,求 分析: 假设,则k mod (i+1) = k - (i+1)*p = k - i*p - p = k mod i - p 则对于某个区间,i∈[l, r],k/i的整数部分p相同,则其余数成等差数列,公差为-p 然后我想到了做莫比乌斯反演时候有个分块加速,在区间[i, n / (n / i)],n/i的整数部分相同,于是有了这份代码. #include <cstdio> #include <algorithm> using namespace std;…

/** 题目:Joseph's Problem 链接:https://vjudge.net/problem/UVA-1363 题意:给定n,k;求k%[1,n]的和. 思路: 没想出来,看了lrj的想法才明白. 我一开始往素数筛那种类似做法想. 想k%[1,n]的结果会有很多重复的,来想办法优化. 但没走通. 果然要往深处想. 通过观察数据发现有等差数列.直接观察很难确定具体规律:此处应该想到用式子往这个方向推导试一试. lrj想法: 设:p = k/i; 则:k%i = k-i*p; 容易想到…

链接:https://vjudge.net/problem/UVA-1363 题意:给出n  k,当 i 属于 1~n 时 ,求解 n% i 的和 n 和 k 的范围都是 1 到 10^9; 商相同 的余数数列 是 公差为商 的 递减等差数列 应该让k / i相等的一连串k % i相加,举个例子: 100 / 34 = 2 ... 32 100 / 35 = 2 ... 30 100 / 36 = 2 ... 28 ... 100 / 50 = 2 ... 0 递减等差数列通项公式:an=a1-…

链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4109 题意: 输入正整数n和k(1≤n,k≤1e9),计算sum(k mod i)(1≤i≤n). 分析: 被除数固定,除数逐次加1,直观上余数也应该有规律.假设k/i的整数部分等于d,则k mod i = k-i*d.因为k/(i+1)和k/i差别不大,如果k/(i+1)的整数部…

https://vjudge.net/problem/UVA-1363 n 题意:求 Σ  k%i i=1 除法分块 如果 k/i==k/(i+1)=p 那么 k%(i+1)=k-(i+1)*p= k-i*p-p = k%i-p 所以 商相同时,余数为等差数列 我不知道为什么交到vjudge一直WA,网上搜的题解交上去也WA #include<cmath> #include<cstdio> using namespace std; int main() { int n,k,i,j,…

题意:给你正整数n和k,然后计算从i到n k%i的和: 思路:如果n小于1000000,直接暴力计算,然后大于1000000的情况,然后在讨论n和k的大小,根据k%i的情况,你会发现规律,是多个等差数列,然后你把这些等差数列加上就是答案. #include <cstdio> #include <cstring> #include <algorithm> #define ll long long using namespace std; ll n,k; ll Getsum…

题意:给定n, k,求出∑ni=1(k mod i) 思路:由于n和k都很大,直接暴力是行不通的,然后在纸上画了一些情况,就发现其实对于k/i相同的那些项是形成等差数列的,于是就可以把整个序列进行拆分成[k,k/2],[k/2, k/3], [k/3,k/4]...k[k/a, k/b]这样的等差数列,利用大步小步算法思想,这里a枚举到sqrt(k)就可以了,这样就还剩下[1,k/a]的序列需要去枚举,总时间复杂度为O(sqrt(k)),然后注意对于n大于k的情况,n超过k的部分全是等于k,为(…

给定$n,k$$(1\leqslant n,k\leqslant 10^9)$,计算$\sum\limits _{i=1}^nk\: mod\:i$ 通过观察易发现$k\%i=k-\left \lfloor \frac{k}{i} \right \rfloor*i$,因此我们考虑把$\left \lfloor \frac{k}{i} \right \rfloor$的值相同的$i$分成一组直接求和,复杂度为$O(\sqrt{n})$. 整除分块原理(选自某dalao博客) #include<bit…

题意:给出n和k,1≤n,k≤1e9,计算 切入点是k/i 和 k/(i+1)差距不大.令pi = k/i, ri = k%i.如果pi+1 == pi,那么ri+1 == k - pi(i+1) == ri - pi, 对于pi+z == pi,ri+z == ri - z*pi,这是等差数列可以O(1)计算出来.如果上界为j,那么k/j ≤ pi,j ≤ k/pi. /********************************************************* * --…

把整个序列进行拆分成[k,k/2),[k/2, k/3), [k/3,k/4)...k[k/a, k/b)的形式,对于k/i(整除)相同的项,k%i成等差数列. /*by SilverN*/ #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; long long n,m,k; i…

题意:给定 n,k,求 while(i <=n) k % i的和. 析:很明显是一个数论题,写几个样例你会发现规律,假设 p = k / i.那么k  mod i = k - p*i,如果 k / (i+1) 也是p,那么就能得到 : k mod (i+1) = k - p*(i+1) = k mod i - p.所以我们就能得到一个等差数列 k mod (i+1) - k mod i = -p,首项是 p % i. 代码如下: #pragma comment(linker, "/STAC…

传送门 题意 计算 \(\sum_{i=1}^n(kmodi)\) 分析 1.n>k 直接输出k*(n-k) 2.n<=k 我们发现k/i相同的k%i构成一个等差数列,那么我们从k/i->2枚举,计算等差数列,最后处理一个[1,sqrt(k))的区间数就好了 复杂度:\(2*O(sqrt(k))\) trick 代码 #include<cstdio> #include<cstring> #include<algorithm> using namespa…

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1512    Accepted Submission(s): 948 Problem Description   The Joseph's problem is notoriously known. For those who are not familiar with th…

Joseph Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 44650   Accepted: 16837 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n,…

Joseph Time Limit: 1 Sec  Memory Limit: 64 MB Submit: 493  Solved: 311 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle e…

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2240    Accepted Submission(s): 1361 Problem Description The Joseph's problem is notoriously known. For those who are not familiar with the…

  Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 50596   Accepted: 19239 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, stan…

Joseph Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 47657   Accepted: 17949 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n,…

 Joseph  The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, ..., n, standing in circle every mth is going to be executed and only the life of the last remaining per…

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1652    Accepted Submission(s): 1031 Problem Description The Joseph's problem is notoriously known. For those who are not familiar with th…

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2094    Accepted Submission(s): 1273 Problem Description The Joseph's problem is notoriously known. For those who are not familiar with the…

一:首先科普一下约瑟夫问题的数学方法 (1)  不管是用list实现还是用vector实现都有一个共同点:要模拟整个游戏过程,不仅程序写起来比較烦,并且时间复杂度高达O(nm),当n,m很大(比如上百万,上千万)的时候,差点儿是没有办法在短时间内出结果的.我们注意到原问题不过要求出最后的胜利者的序号,而不是要读者模拟整个过程.因此假设要追求效率,就要打破常规,实施一点数学策略. (2) 为了讨论方便,先把问题略微改变一下,并不影响原意:  问题描写叙述:n个人(编号0~(n-1)),从0開始报数…

题目链接: POJ  1012: id=1012">http://poj.org/problem?id=1012 HDU 1443: pid=1443">http://acm.hdu.edu.cn/showproblem.php? pid=1443 约瑟夫环(百度百科): fr=aladdin" target="_blank">http://baike.baidu.com/view/717633.htm?fr=aladdin Descri…

Joseph Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 50068   Accepted: 19020 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n,…

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2126    Accepted Submission(s): 1291 Problem Description The Joseph's problem is notoriously known. For those who are not familiar with th…

题意: 给n,m,和m个数(k1~km).求1~n中有多少个数不是(k1~km)中任意一数的倍数. 题解: 容斥模板题.反面考虑,a的倍数有n/a个:既是a,也是b的倍数,即lcm(a,b)的倍数有n/lcm(a,b)个.是a,b,c的倍数,即lcm(a,b,c)的倍数有n/lcm(a,b,c)个. #include<iostream> #include<cstdio> #include<algorithm> using namespace std; typedef l…

Joseph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2453    Accepted Submission(s): 1476 Problem Description The Joseph's problem is notoriously known. For those who are not familiar with the…

Joseph Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 52097   Accepted: 19838 Description The Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n,…

The Sports Association of Bangladesh is in great problem with their latest lottery `Jodi laiga Jai'. Thereare so many participants this time that they cannot manage all the numbers. In an urgent meeting theyhave decided that they will ignore some num…

1. 几何 4 1.1 注意 4 1.2 几何公式 4 1.3 多边形 6 1.4 多边形切割 9 1.5 浮点函数 10 1.6 面积 15 1.7 球面 16 1.8 三角形 17 1.9 三维几何 19 1.10 凸包 26 1.11 网格 28 1.12 圆 28 1.13 整数函数 30 2. 组合 33 2.1 组合公式 33 2.2 排列组合生成 33 2.3 生成gray码 35 2.4 置换(polya) 35 2.5 字典序全排列 36 2.6 字典序组合 36 3. 结构…