#include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<vector> #define MAXN 100000+10 #define ll long long #define pb push_back #define ft first #define sc second #define mp make_pair #define pil pa…

http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; int jc[100003]; int p; int ipow(int x, int b) { ll t = 1, w = x;…

Problem Description Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long cold days. After some time the squirrel family thinks that they have to solve a problem. They supp…

http://acm.hdu.edu.cn/showproblem.php?pid=5025 Saving Tang Monk Problem Description   <Journey to the West>(also <Monkey>) is one of the Four Great Classical Novels of Chinese literature. It was written by Wu Cheng'en during the Ming Dynasty.…

Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14342    Accepted Submission(s): 6420 Problem Description 话说上回讲到海东集团面临内外交困,公司的元老也只剩下XHD夫妇二人了.显然,作为多年拼搏的商人,XHD不会坐以待毙的.  一天,当他正在苦思冥想解困良策的…

排列组合 啊……这题是要求c(n-1,0)+c(n,1)+c(n+1,2)+......+c(n+m-1,m) 这个玩意……其实就等于c(n+m,m) 好吧然后就是模P……Lucas大法好= = 我SB地去预处理<P的所有fac和inv了……果断TLE 事实上Lucas时对于<P的部分直接暴力算就好了 //HDOJ 3037 #include<cstdio> #include<cstdlib> #include<cstring> #include<io…

hdu 3037 Saving Beans 题目大意:n个数,和不大于m的情况,结果模掉p,p保证为素数. 解题思路:隔板法,C(nn+m)多选的一块保证了n个数的和小于等于m.可是n,m非常大,所以用到Lucas定理. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; ll n, m, p; ll qPow (ll a…

Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8258    Accepted Submission(s): 3302 Problem Description Although winter is far away, squirrels have to work day and night to save be…

Saving Beans Time Limit: 3000 MS Memory Limit: 32768 K Problem Description Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long cold days. After some time the squirrel fam…

Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4315    Accepted Submission(s): 1687 Problem Description Although winter is far away, squirrels have to work day and night to save be…

Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2079    Accepted Submission(s): 748 Problem Description Although winter is far away, squirrels have to work day and night to save bea…

Saving Beans Time Limit: 3000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 303764-bit integer IO format: %I64d      Java class name: Main   Although winter is far away, squirrels have to work day and night to save beans. T…

如果您有n+1树,文章n+1埋不足一棵树m种子,法国隔C[n+m][m] 大量的组合,以取mod使用Lucas定理: Lucas(n,m,p) = C[n%p][m%p] × Lucas(n/p,m/p,p) ; Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2314    Accepted Submissi…

Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5769    Accepted Submission(s): 2316 Problem Description Although winter is far away, squirrels have to work day and night to save b…

Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long c…

hdu3037 Saving Beans 题意:n个不同的盒子,每个盒子里放一些球(可不放),总球数<=m,求方案数. $1<=n,m<=1e9,1<p<1e5,p∈prime$ 卢卡斯(Lucas)定理(计算组合数 防爆精度) $lucas(n,m,p)=lucas(n/p,m/p,p)*C(n\%p,m\%p,p)$ $lucas(n,0,p)=1$ 老套路,插板法. 设m个球都要放,多一个盒子轻松解决. 根据插板法得方案数$=C(n+m,n)$ 跑一遍Lucas. en…

Saving Beans Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4153    Accepted Submission(s): 1607 Problem Description Although winter is far away, squirrels have to work day and night to save be…

链接:HDU - 6409:没有兄弟的舞会 题意: 题解: 求出最大的 l[i] 的最大值 L 和 r[i] 的最大值 R,那么 h 一定在 [L, R] 中.枚举每一个最大值,那么每一个区间的对于答案的贡献就是一个等差数列的和(乘法分配律),将每一个和乘起来就是该最大值的对于答案的贡献.但是相同最大值可能来自于多个区间,如果枚举每一个可能出现最大值的区间,那么会超时,所以需要一个神奇的方法,我也不知道原理是什么,但是数学上就是直接的式子,可以分解出来. #include <bits/stdc+…

EJB EJB定义了一组可重用的组件:Enterprise Beans.开发人员可以利用这些组件,像搭积木一样建立分布式应用.…

http://poj.org/problem?id=3321 http://acm.hdu.edu.cn/showproblem.php?pid=3887 POJ 3321: 题意:给出一棵根节点为1的边不一定的树,然后给出问题:询问区间和 或者 节点值更新. HDU 3887: 题意:和POJ 3321的题意差不多,只不过对每个节点询问不包含该节点的区间和 思路:今天才学了下才知道有DFS序这种东西,加上树状数组处理一下区间和 和 节点更新. DFS序大概就是我们在DFS遍历一棵树的时候,在进…

主题链接:pid=3037">http://acm.hdu.edu.cn/showproblem.php?pid=3037 推出公式为C(n + m, m) % p. 用Lucas定理求大组合数取模的值 代码: #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; int t; long long n, m, p; long long pow(lo…

Problem Description Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long cold days. After some time the squirrel family thinks that they have to solve a problem. They supp…

解题思路: 直接求C(n+m , m) % p , 由于n , m ,p都非常大,所以要用Lucas定理来解决大组合数取模的问题. #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string&g…

题意:问用不超过 m 颗种子放到 n 棵树中,有多少种方法. 析:题意可以转化为 x1 + x2 + .. + xn = m,有多少种解,然后运用组合的知识就能得到答案就是 C(n+m, m). 然后就求这个值,直接求肯定不好求,所以我们可以运用Lucas定理,来分解这个组合数,也就是Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p). 然后再根据费马小定理就能做了. 代码如下: 第一种: #pragma comment(linker, "/STACK:10240…

acm.hdu.edu.cn/showproblem.php?pid=3037 [题意] m个松果,n棵树 求把最多m个松果分配到最多n棵树的方案数 方案数有可能很大,模素数p 1 <= n, m <= 1000000000, 1 < p < 100000 [思路] 答案为C(n+m,m)%p 对于C(n, m) mod p.这里的n,m,p(p为素数)都很大的情况.就不能再用C(n, m) = C(n - 1,m) + C(n - 1, m - 1)的公式递推了.这里用到Luca…

原题直通车:HDU 4308 Saving Princess claire_ 分析: 两次BFS分别找出‘Y’.‘C’到达最近的‘P’的最小消耗.再算出‘Y’到‘C’的最小消耗,比较出最小值 代码: #include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<string> using namespace std; const int inf=0xF…

http://acm.hdu.edu.cn/showproblem.php?pid=2732 题意:给出两个地图,蜥蜴从一个柱子跳跃到另外一个地方,那么这个柱子就可能会坍塌,第一个地图是柱子可以容忍跳跃多少次(即从它为起点可以跳跃多少次,然后坍塌),第二个地图是蜥蜴的位置.还有一个跳跃距离d,即蜥蜴可以跳跃不超过d的距离(曼哈顿距离),如果跳到地图外面,即蜥蜴逃跑成功,问最终留下的蜥蜴数最少是多少. 思路:我觉得如果不是在做网络流套题应该想不到这是网络流.其实挺简单的建图.首先考虑蜥蜴的位置,和…

http://acm.hdu.edu.cn/showproblem.php?pid=4289 题意:有n个城市,m条无向边,小偷要从s点开始逃到d点,在每个城市安放监控的花费是sa[i],问最小花费可以监控到所有小偷. 思路:求最小割可以转化为最大流.每个城市之间拆点,流量是sa[i],再增加一个超级源点S和s相连,增加一个超级汇点T,让d的第二个点和T相连.然后就可以做了. #include <cstdio> #include <algorithm> #include <i…

http://acm.hdu.edu.cn/showproblem.php?pid=3605 题意:有n个人要去到m个星球上,这n个人每个人对m个星球有一个选择,即愿不愿意去,"Y"or"N".问是否可以全部人都顺利到自己想去的星球. 思路:很"有趣"的一道题目,n是1e5的大小,m只有10,没有想到状态压缩,看到n这么大肯定超时还是强行写了一波,于是RE(TLE).想了挺久还是不会.看别人的思路是说二进制状态压缩.看到这就想到m只有10,于是可…

http://acm.hdu.edu.cn/showproblem.php?pid=4292 题意:和奶牛一题差不多,只不过每种食物可以有多种. 思路:因为食物多种,所以源点和汇点的容量要改下.还有Dinic又TLE了,用ISAP过. #include <cstdio> #include <algorithm> #include <iostream> #include <cstring> #include <string> #include &l…