HDU1028Ignatius and the Princess III(一个数有多少种组合方式,DP)与放n个苹果在m个盘子有多少生种组合一样(两种方法解)
2015-06-03 14:33
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Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15210 Accepted Submission(s): 10724
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];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"
Input
The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.
Output
For each test case, you have to output a line contains an integer P which indicate the different equations you have found.
Sample Input
4 10 20
Sample Output
5 42 627
Author
Ignatius.L
解题:把n看作是苹果的数量,每个位置看作是盘子,那么最多有n个盘子。第m个盘子最多放n/m个,也就是每个盘放苹果数量服从的原则是:非递增式放数量。
#include<stdio.h> #include<string.h> const int N = 125; #define LL __int64 LL dp ,sum ; int main(){ memset(dp,0,sizeof(dp)); memset(sum,0,sizeof(sum)); for(int i=1;i<=124;i++) dp[i][1][i]=1; sum[1][1]=1; for(int i=2;i<=124;i++)//苹果数 for(int j=2;j<=i;j++){ //盘子数 for(int k=0;k<=i/j;k++){ //最后一个盘放的数量 dp[i][j][k]=0; for(int tk=k;tk<=(i-k)/(j-1);tk++)//倒数第二个盘放的数量 dp[i][j][k]+=dp[i-k][j-1][tk]; sum[i][j]+=dp[i][j][k]; } } int n; while(scanf("%d",&n)>0) printf("%I64d\n",sum ); }
方法二:方法出处
#include<stdio.h> #include<string.h> const int N = 125; #define LL __int64 LL sum ; int main(){ memset(sum,0,sizeof(sum)); for(int i=1;i<=124;i++) sum[1][i]=sum[i][1]=1; for(int i=2;i<=120;i++) for(int j=2;j<=120;j++){ if(j>i)sum[i][j]=sum[i][i]; else if(i==j)sum[i][j]=sum[i][j-1]+1; else sum[i][j]=sum[i][j-1]+sum[i-j][j]; } int n; while(scanf("%d",&n)>0) printf("%I64d\n",sum ); }
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