题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=2110 题意:给你N个价值和数目,求方案数,很裸的母函数. #include<cstdio> ],c2[],all,pi[],mi[],su,i,j,k; int mu(int sum){ ;i<=sum;i++)c1[i]=,c2[i]=; ;i<=mi[];i++)c1[pi[]*i]=; ;i<=all;i++){ ;j<=mi[i];j++) ;pi[i]*j+k&l…

Big Event in HDU DescriptionNowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.The splitting is absolute…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=2110 题意:分出1/3的价值,有几种可能? 思路:母函数 与之前的题目2079相似,复习笔记再来写代码: https://www.cnblogs.com/CheeseIce/p/9595315.html 代码: #include <iostream> #include <cstring> using namespace std; #define min(a,b) ((a)<(b)?(…

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002. The splitting is absolutely a big event in HDU! At t…

欢迎参加——BestCoder周年纪念赛(高质量题目+多重奖励) Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28468    Accepted Submission(s): 10023 Problem Description Nowadays, we all know that Computer…

Crisis of HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3606    Accepted Submission(s): 1015 Problem Description 话说上回讲到HDU大战东洋小苟,结果自然是中方大胜,这一战也使得海东集团在全球同行业中的地位更加巩固.随着集团的发展,非常多创业时期的元老逐步功成身…

Big Event in HDU  HDU1171 就是求一个简单的背包: 题意:就是给出一系列数,求把他们尽可能分成均匀的两堆 如:2 10 1 20 1     结果是:20 10.才最均匀! 三种解法: 多重背包的优化与否:(1031MS) #include<iostream> #include<stdio.h> #include<algorithm> #include<string.h> using namespace std; ]; ],b[];…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24002    Accepted Submission(s): 8458 Problem Description Nowadays, we all know that Computer College is the biggest department…

找单词 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4093    Accepted Submission(s): 2933 Problem Description 假设有x1个字母A, x2个字母B,..... x26个字母Z,同时假设字母A的价值为1,字母B的价值为2,..... 字母Z的价值为26.那么,对于给定的字母,可以找到…

这种有限制的类棋盘着色问题一般可以用指数型母函数来解决,设Hn表示这样的着色数,首先H0=1,则Hn等于四个字母的(A,B,C,D)的多重集合的n排列数,其中每个字母的重数是无穷,且要求A,C出现的次数是偶数,因此,H0,H1,...Hn,...的指数生成函数是A,B,C,D因子的乘积: 用快速幂解决,只不过在HDU不能用long long解决,要用__int64. 代码: #include <iostream> #include <cstdio> #include <cst…