Devu and Birthday Celebration 我们发现不合法的整除因子在 m 的因子里面, 然后枚举m的因子暴力容斥, 或者用莫比乌斯系数容斥. #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pa…
这是本人第一次写代码,难免有点瑕疵还请见谅 A. Devu, the Singer and Churu, the Joker time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Devu is a renowned classical singer. He is invited to many big functions/festi…
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…
题意:给出a数组和b数组,他们的长度最大1e5,元素范围是1到1e9,问你让a数组最小的数比b数组最大的数要大需要的最少改变次数是多少.每次改变可以让一个数加一或减一 分析:枚举a数组和b数组的所有的元素x,作为他们的界限,也就是说a数组所有的数要大于等于x,b数组所有的数要小于等于x,再利用前缀和+二分,分别求出ab数组需要改变的次数,在所有的方案中取一个最小值 代码: #include <bits/stdc++.h> #define ll long long using namespace…
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…
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…
D. Devu and his Brother time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Devu and his brother love each other a lot. As they are super geeks, they only like to play with arrays. They are giv…
C. Devu and Partitioning of the Array time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he ca…
