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…

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…

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…

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): 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): 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) 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 /*******…