---恢复内容开始--- 2017-08-10 20:00:45 writer:pprp 拆分数: 把正整数n拆分成k个正整数之和的方案数: 问题转换:将1转化为2 1.把n表示成m个正整数之和的方案数 2.把n表示成不超过m的正整数之和的方案数 两者答案相同:解释Ferrers图 用dp来做,dp[i][j]的意思是 i 表示成 不超过 j 的方案数 边界条件: dp[0][j] = 1 , dp[[i][1] = 1 , dp[1][j] = 1; 状态转移: dp[i][j] = dp[i…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15942    Accepted Submission(s): 11245 Problem Description "Well, it seems the first problem is too easy. I will let…

以下引用部分全都来自:http://blog.csdn.net/ice_crazy/article/details/7478802  Ice—Crazy的专栏 分析: HDU 1028 摘: 本题的意思是:整数划分问题是将一个正整数n拆成一组数连加并等于n的形式,且这组数中的最大加数不大于n. 如6的整数划分为 6 5 + 1 4 + 2, 4 + 1 + 1 3 + 3, 3 + 2 + 1, 3 + 1 + 1 + 1 2 + 2 + 2, 2 + 2 + 1 + 1, 2 + 1 + 1…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11810    Accepted Submission(s): 8362 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III HDU - 1028 整数划分问题 假的dp(复杂度不对) #include<cstdio> #include<cstring> typedef long long LL; LL ans[][]; LL n,anss; LL get(LL x,LL y) { ) return ans[x][y]; ) ; ; ans[x][y]=; LL i; ;i<=y;i++) ans[x][y]+=get(x-y,i); re…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15498    Accepted Submission(s): 10926 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10312    Accepted Submission(s): 7318 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says. &…

题目链接:hdu 1028 Ignatius and the Princess III 题意:对于给定的n,问有多少种组成方式 思路:dp[i][j],i表示要求的数,j表示组成i的最大值,最后答案是dp[i][i].那么dp[i][j]=dp[i][j-1]+dp[i-j][i-j],dp[i][j-1]是累加1到j-1的结果,dp[i-j][i-j]表示的就是最大为j,然后i-j有多少种表达方式啦.因为i-j可能大于j,这与我们定义的j为最大值矛盾,所以要去掉大于j的那些值 /*******…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12521    Accepted Submission(s): 8838 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24975    Accepted Submission(s): 17253 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9532    Accepted Submission(s): 6722 Problem Description "Well, it seems the first problem is too easy. I will let y…

Ignatius and the Princess III Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 56   Accepted Submission(s) : 41 Problem Description "Well, it seems the first problem is too easy. I will let you kn…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16028    Accepted Submission(s): 11302 Problem Description "Well, it seems the first problem is too easy. I will let…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1028 Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24967    Accepted Submission(s): 17245 Problem Description "Well…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25805    Accepted Submission(s): 17839 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16122    Accepted Submission(s): 11371 Problem Description "Well, it seems the first problem is too easy. I will le…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13553    Accepted Submission(s): 9590 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25929    Accepted Submission(s): 17918 Problem Description "Well, it seems the first problem is too easy. I will let…

Ignatius and the Princess III Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 26219    Accepted Submission(s): 18101 Problem Description "Well, it seems the first problem is too easy. I will let…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1028 思路分析:该问题要求求出某个整数能够被划分为多少个整数之和(如 4 = 2 + 2, 4 = 2 + 1 + 1),且划分的序列 2, 2 或者 2, 1, 1为单调非递增序列: 使用动态规划解法:假设dp[i][j]表示整数i被划分的序列中最大值不超过j的所有的可能,则使用类似于0-1背包的思考方法: (1)如果i == j,则dp[i][j] = dp[i][j-1] + 1: (2)如果…

http://acm.hdu.edu.cn/showproblem.php?pid=1028 dp[i][j]表示数值为i,然后最小拆分的那个数是j的时候的总和. 1 = 1 2 = 1 + 1 .  2 = 2 3 = 1 + 1 + 1. 3 = 2 + 1. 3 = 3 那么可以分两类, 1.最小拆分数是j,这个时候dp[i][j] = dp[i - j][j].加一个数j,使得它变成i 2.最小拆分数严格大于j,这个时候就没得加上j了.就是dp[i][j + 1] 所以dp[i][j]…

Problem Description "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says."The second problem is, given an positive integer N, we define an equation like this:  N=a[1]+a[2]+a[3]+...+a[m…

整数的有序拆分就是隔板法,无序拆分则有两种处理方法 DP递推 我们假设P(n,m)P(n,m)P(n,m)是正整数nnn无序拆分为mmm个正整数的方案数 对于某一种拆分,不妨将拆分出来的mmm个数从小到大排序,分类讨论 最小的数等于111,那么去掉这个111,相当于把剩下的n−1n-1n−1拆分成m−1m-1m−1个数,方案数就为P(n−1,m−1)P(n-1,m-1)P(n−1,m−1) 最下的数大于111,那么将所有的数减去111,相当于把剩下的n−mn-mn−m拆分成mmm个数,方案数就为…

题目链接:HDU 1028 Problem Description "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says. "The second problem is, given an positive integer N, we define an equation like this: N=a[1]+a[2…

题目: "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says. "The second problem is, given an positive integer N, we define an equation like this:   N=a[1]+a[2]+a[3]+...+a[m];   a[i]>0…

链接:传送门 题意:一个数n有多少种拆分方法 思路:典型母函数在整数拆分上的应用 /************************************************************************* > File Name: 1.cpp > Author: WArobot > Blog: http://www.cnblogs.com/WArobot/ > Created Time: 2017年04月20日 星期四 21时07分09秒 ********…

本题应该有两种方法: 1.母函数法 2.递推法 母函数不了解,待充分了解之后,再进行补充! 这里为递推实现的方法: 思路: 定义:n为要拆分的整数: k为拆分的项数: f[n][k]代表 n的整数拆分中,最大项不超过k的方案数. 每一个整数n的拆分中,总有一项拆分为自己,即:n = n; 我们将其表示为f[n][1],而且f[n][1] = 1; 又,每一个整数n的拆分中,总有一项拆分为n个1,即:n = 1 + 1 + ...... + 1(n个1的加和); 我们将其表示为f[0][0],且f…

题目传送门:https://vjudge.net/problem/HDU-1028 思路:整数拆分构造母函数的模板题 1 //#include<bits/stdc++.h> 2 #include<time.h> 3 #include <set> 4 #include <map> 5 #include <stack> 6 #include <cmath> 7 #include <queue> 8 #include <c…

Description "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says. "The second problem is, given an positive integer N, we define an equation like this:   N=a[1]+a[2]+a[3]+...+a[m];   a…