HDU 2152 Fruit( DP )】的更多相关文章

Fruit Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3514    Accepted Submission(s): 1998 Problem Description 转眼到了收获的季节,由于有TT的专业指导,Lele获得了大丰收.特别是水果,Lele一共种了N种水果,有苹果,梨子,香蕉,西瓜……不但味道好吃,样子更是好看. 于是,…
系数为1的母函数…… #include <cstdio> #include <cstring> using namespace std; int n,m,size[105][2],c1[105],c2[105]; int main(){ while(scanf("%d%d",&n,&m)!=EOF){ for(int i=1;i<=n;i++)scanf("%d%d",&size[i][0],&size[…
# include<stdio.h> # include <algorithm> # include <string.h> # include <iostream> using namespace std; int main() { int n,m,i,j,k; int c1[110],c2[110]; int a[110],b[110]; while(~scanf("%d%d",&n,&m)) { for(i=0;i&l…
HDU - 2152 思路 :有上下界的生成函数 .生成函数即母函数,是组合数学中尤其是计数方面的一个重要理论和工具.生成函数有普通型生成函数和指数型生成函数两种, 其中普通型用的比较多.形式上说,普通型生成函数用于解决多重集的组合问题,此处 方案 不同的定义为如果各种水果的数目都相同,则认为这两种方案是相同的.所以为 多重集的组合问题. #include<bits/stdc++.h> using namespace std; #define maxn 123 struct node { in…
http://acm.hdu.edu.cn/showproblem.php? pid=4123 Problem Description Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads in his village. Each road connects two houses,…
http://acm.hdu.edu.cn/showproblem.php?pid=4507 Problem Description 单身! 依旧单身! 吉哥依旧单身! DS级码农吉哥依旧单身! 所以.他生平最恨情人节,无论是214还是77.他都讨厌! 吉哥观察了214和77这两个数,发现: 2+1+4=7 7+7=7*2 77=7*11 终于,他发现原来这一切归根究竟都是由于和7有关!所以,他如今甚至讨厌一切和7有关的数. 什么样的数和7有关呢? 假设一个整数符合以下3个条件之中的一个.那么我…
http://acm.hdu.edu.cn/showproblem.php?pid=3709 Problem Description A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit.…
XHXJ's LIS Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2265    Accepted Submission(s): 927 Problem Description #define xhxj (Xin Hang senior sister(学姐)) If you do not know xhxj, then careful…
You Are the One Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3348    Accepted Submission(s): 1524 Problem Description The TV shows such as You Are the One has been very popular. In order to m…
HDU 2829 区间DP & 前缀和优化 & 四边形不等式优化 n个节点n-1条线性边,炸掉M条边也就是分为m+1个区间 问你各个区间的总策略值最少的炸法 就题目本身而言,中规中矩的区间DP问题 d p[i][j]表示前i个节点,分为j个区间的最优策略值 cost[i][j]为从i到j节点的策略值 所以dp[i][j] = min(dp[k-1][j-1] + cost[k][i] 但是复杂度太高了 可以优化的地方有: cost数组值得求取: 考虑到cost(i,j)=ΣAxAy (i≤…