题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1028 题意: 给你一个正整数n,将n拆分成若干个正整数之和,问你有多少种方案. 注:"4 = 3 + 1"和"4 = 1 + 3"视为同一种方案.(加数是无序的) 题解1(dp): 表示状态: dp[n][m] = num of methods 表示用均不超过m的元素组成n的方法数. 如何转移: 假设当前状态为dp[n][m]. 对于等于m的元素,有两种决策.要么不用,…

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…

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): 11810    Accepted Submission(s): 8362 Problem Description "Well, it seems the first problem is too easy. I will let…

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 /*******…

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 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) Problem Description "Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says. &…

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: 2000/1000 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): 15942    Accepted Submission(s): 11245 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): 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…

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): 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 : 2000/1000ms (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…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1028 题目大意:3=1+1+1=1+2=3 :4=4=1+1+1+1=1+2+1=1+3:所以3有3种加法,4有4种加法,给出一个n,(1<=n<=120) ,计算n有几种加法. 解法:母函数的最简单的一道入门题之一. 说明:今早上突发奇想着想学学母函数,就搜了搜HDU ACM PPT 的母函数一版,把这道题分享给大家. 感想:这道题之前看过,推过公式什么的,没推出来.今天才知道是母函数的算法.算…

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…

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…

HDU - 1028 step 1:初始化第一个多项式 也就是 由 1的各种方案 组 成 的多项式 初始化系数为 1.临时区 temp初始化 为 0 step 2:遍历后续的n - 1 个 多项式 ,第二重 for  j  代 表 的 存 储 结 果 的 多 项 式的次数,k 代表 当前 第 i 的 多项式的次数 通过计算发现两个多项式相乘 其中一个 系数为1和 0 组成,运算时可以初始化系数数组为0 ,然后 由另一个的系数 与之相加即可得到 G(x)=(1+x+x2+x3+x4+.....)(…

题意: 输入一个数n,求组合成此数字可以有多少种方法,每一方法是不记录排列顺序的.用来组成的数字可以有1.2.3....n.比如n个1组成了n,一个n也组成n.这就算两种.1=1,2=1+1=2,3=3=1+2=1+1+1,而1+2和2+1只能算一种.n最大为120. 思路:关于母函数的原理不讲了.讲怎么实现几个括号相乘. 思路: 我们要算的n是等于120,把其简化为5,就是说设n最大为5,道理一样的.5一共有7种方法对吗!自己手写吧. 如果想要得出结果,那么一共有5个括号要相乘,分别如下: 为…

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]…

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…

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];…

简单的钱币兑换问题,就是钱的种类多了一点,完全背包. #include<cstdio> #include<cstring> int main () { ]; memset(dp,,sizeof(dp)); dp[]=; ; i<=; i++) ; j++) dp[j]+=dp[j-i]; while(~scanf("%d",&i)) printf("%d\n",dp[i]); ; }…