hdu1028 Ignatius and the Princess III（DP整数划分）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15663    Accepted Submission(s): 11042

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

41020

 

Sample Output

542627

 

题意：给定一个整数，将其写成若干个整数的和，求有多少种情况。

可以直接考虑各个加数递减的情况，避免重复。依旧是DP的思想。

将整数n划分成两个整数之和，n+0，(n-1)+1，(n-2)+2，……，(n-n/2)+(n/2)。再将加号前面的加数继续如此划分，注意：继续划分时，后一个加数至少为上一级的后一个加数，以确保加数列的递减性。用a[n][1]表示将n表示成(n-1)+1的情况数，则可知a[n][1]=a[n-1][0]+a[n-1][1]+a[n-1][2]+……+a[n-1][(n-1)/2]+1。

a[i][j]=a[i-1][0]+a[i-1][j]+a[i-1][j+1]+……+a[i-1][(i-1)/2]。把a[i][0]都初始化为1，表示不划分时就一种情况。最后，将所有a[i][j]都加到a[i][0]上，表示i的所有划分情况数。

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;int a[125][70];//a[i][j]表示和为i 且 两个加数的后一个为j的情况数量（j<=i/2）void biao(){    for(int i = 0; i < 125; i++)    {        a[i][0]=1;        for(int j = 1; j <= i / 2; j++)        {            for(int k = j; k <= (i - j) / 2; k++)                a[i][j] += a[i-j][k];            a[i][j]++;            a[i][0] += a[i][j];        }    }}int main(){    //freopen("in.txt","r",stdin);    biao();    int n;    while (scanf("%d",&n)!=EOF)    {        printf("%d\n",a[n][0]);    }}