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…

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…

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…

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,…

这是道典型的母函数的题目,可以看看我的母函数这一标签上的另一道例题,里面对母函数做了较为详细的总结.这题仅贴上代码: #include"iostream" using namespace std; #define N 130 ],b[N+]; int main() { int n,i,j,k; ) { ;i<=n;i++) {a[i]=;b[i]=;} ;i<=n;i++) { ;j<=n;j++) ;k+j<=n;k+=i) { b[k+j]+=a[j]; }…

题意 给出$n$,问用$1$到$n$的数字问能构成$n$的方案数 思路 生成函数基础题,$x^{n}$的系数即答案. 代码 #include <bits/stdc++.h> #define DBG(x) cerr << #x << " = " << x << endl; using namespace std; const int N = 120 + 5; int n, c[2][N]; int main() { while(…

先咕着 ---------------2018 5 22---------------------- 题解 生成函数处理整数拆分 code #include<cstdio> #include<cstring> #include<algorithm> inline int raed() { int x = 0,f = 1; char c = getchar(); while(c < '0' || c < '9') c = getchar(); while(c…

大意是给你1个整数n,问你能拆成多少种正整数组合.比如4有5种: 4 = 4;  4 = 3 + 1;  4 = 2 + 2;  4 = 2 + 1 + 1;  4 = 1 + 1 + 1 + 1; 然后就是母函数模板题……小于n的正整数每种都有无限多个可以取用. (1+x+x^2+...)(1+x^2+x^4+...)...(1+x^n+...) 答案就是x^n的系数. #include<cstdio> #include<cstring> using namespace std;…

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) 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): 15942    Accepted Submission(s): 11245 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 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): 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): 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): 11810    Accepted Submission(s): 8362 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: 2000/1000 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): 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): 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): 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 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 lik…

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…

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…

---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…