poj 3790 Recursively Palindromic Partitions

/*
摘抄自博客：
Recursively Palindromic Partitions
Time Limit: 1000MS        Memory Limit: 65536K
Total Submissions: 472        Accepted: 345
Description

A partition of a positive integer N is a sequence of integers which sum to N, usually written with plus signs between the numbers of the partition. For example 

15 = 1+2+3+4+5 = 1+2+1+7+1+2+1 

A partition is palindromic if it reads the same forward and backward. The first partition in the example is not palindromic while the second is. If a partition containing m integers is palindromic, its left half is the first floor(m/2) integers and its right half is the last floor(m/2) integers (which must be the reverse of the left half. (floor(x) is the greatest integer less than or equal to x.) 

A partition is recursively palindromic if it is palindromic and its left half is recursively palindromic or empty. Note that every integer has at least two recursively palindromic partitions one consisting of all ones and a second consisting of the integer itself. The second example above is also recursively palindromic. 

For example, the recursively palindromic partitions of 7 are: 

7, 1+5+1, 2+3+2, 1+1+3+1+1, 3+1+3, 1+1+1+1+1+1+1 

Write a program which takes as input an integer N and outputs the number of recursively palindromic partitions of N.
Input

The first line of input contains a single integer N, (1 <= N <= 1000) which is the number of data sets that follow. Each data set consists of a single line of input containing a single positive integer for which the number of recursively palindromic partitions is to be found.
Output

For each data set, you should generate one line of output with the following values: The data set number as a decimal integer (start counting at one), a space and the number of recursively palindromic partitions of the input value.
Sample Input

3 
4 
7 
20
Sample Output

1 4
2 6
3 60
Source

Greater New York Regional 2008


题意是：求每个数分割成递归回文的个数，即将n分割后，再将其左侧和右侧分割，依次..自身也算一种；
以7为列：
0+7+0 f[0]；
1+5+1 f[1];
2+3+2 f[2];
3+1+3 f[3];
f[7]=f[0]+f[1]+f[2]+f[3];
每个数的递归回文个数都等于它一半之前所有回文个数之和；其中f[0]=1;f[1]=1;
递归即可；偶数和其之后的奇数回文个数是相等的；*/
#include<stdio.h>
int f[1005];
int main()
{
    int n,m,i,j;
    f[0]=1;
    f[1]=1;
    for(j=2; j<=1000; j++)
    {
        f[j]=0;
        for(i=0; i<=j/2; i++) f[j]+=f[i];
    }
    scanf("%d",&n);
    for(i=1; i<=n; i++)
    {
        scanf("%d",&m);
        printf("%d %d
",i,f[m]);
    }
    return 0;
}