Codeforces 451 E Devu and Flowers】的更多相关文章

Discription Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hence they are indistinguishable). Also, no two boxes have flowers of the…
传送门 解题思路: 假如只有 s 束花束并且不考虑 f ,那么根据隔板法的可重复的情况时,这里的答案就是 假如说只有一个 f 受到限制,其不合法时一定是取了超过 f 的花束 那么根据组合数,我们仍然可以算出其不合法的解共有: 最后,由于根据容斥,减两遍的东西要加回来,那么含有偶数个 f 的项为正,奇数个时为负. 答案就是: 搜索答案,使用Lucas定理,计算组合数上下约去. 代码: #include<cstdio> #include<cstring> #include<alg…
[题目链接] http://codeforces.com/contest/451/problem/E [算法] 容斥原理 [代码] #include<bits/stdc++.h> using namespace std; typedef long long ll; ; int i,j,n,s,ans,MASK; ll m,t; ll a[],f[]; ]; inline int power(int a,int n) { ,b = a; while (n) { ) res = 1ll * res…
E. Devu and Flowers 题目连接: http://codeforces.com/contest/451/problem/E Description Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hen…
题目链接:Codeforces 451E Devu and Flowers 题目大意:有n个花坛.要选s支花,每一个花坛有f[i]支花.同一个花坛的花颜色同样,不同花坛的花颜色不同,问说能够有多少种组合. 解题思路:2n的状态,枚举说那些花坛的花取超过了,剩下的用C(n−1sum+n−1)隔板法计算个数.注意奇数的位置要用减的.偶数的位置用加的.容斥原理. #include <cstdio> #include <cstring> #include <cmath> #in…
Codeforces Round #258 (Div. 2) Devu and Flowers E. Devu and Flowers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Devu wants to decorate his garden with flowers. He has purchased n boxes…
E. Devu and Flowers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All fl…
CF451E Devu and Flowers 题意: \(Devu\)有\(N\)个盒子,第\(i\)个盒子中有\(c_i\)枝花.同一个盒子内的花颜色相同,不同盒子的花颜色不同.\(Devu\)要从中选出\(M\)枝花,求有多少种方案,对\(10e9+7\)取模. 数据范围 \(1 \le N \le 20,0 \le M \le 10^{14},0 \le c_i \le 10^{12}\) 其实就是求多重集组合数的模板题. 可以看看我写的博客 一些注意点,发现直接求会爆\(long lo…
CF451E Devu and Flowers(容斥) 题目大意 \(n\)种花每种\(f_i\)个,求选出\(s\)朵花的方案.不一定每种花都要选到. \(n\le 20\) 解法 利用可重组合的公式. 不考虑\(f_i\)的限制,直接可重组合的方案是,意思是从可以重复的\(n\)个元素中取出\(r\)个的个数.注意,根据定义,此时\(r\)种每个都要选. \[ f(s,r)={s+r-1 \choose r-1} \] 考虑限制怎么办,我们先容斥. 我们可以钦定某些花选择了\(f_i+1\)…
Zap FGD正在破解一段密码,他需要回答很多类似的问题:对于给定的整数a,b和d,有多少正整数对x,y,满足x<=a,y<=b,并且gcd(x,y)=d.作为FGD的同学,FGD希望得到你的帮助. Input 第一行包含一个正整数n,表示一共有n组询问.(1<=n<= 50000)接下来n行,每行表示一个询问,每行三个 正整数,分别为a,b,d.(1<=d<=a,b<=50000) Output 对于每组询问,输出到输出文件zap.out一个正整数,表示满足条件…