JOJ2185: Adding 1s, 2s, and 3s

本文介绍了一个简单的递归算法,该算法用于计算一个指定正整数可以如何被拆分为1、2和3的和的不同方式的数量。通过示例输入输出展示了算法的实际运行效果。

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Integer 4 can be expressed as a sum of 1s, 2s, and 3s in seven different ways as follows:
                       1 + 1 + 1 + 1;     (1)
                         1 + 1 + 2;       (2)
                         1 + 2 + 1;       (3)
                         2 + 1 + 1;       (4)
                           2 + 2;         (5)
                           1 + 3;         (6)
                           3 + 1:         (7)
Write a program that determines the number of ways in which a given integer can be expressed as a sum of 1s, 2s, and 3s. You may assume that the integer is positive and less than 20.

Input

The input consists of T test cases. The number of test cases (T ) is given in the first line of the input. Each test case consists of an integer written in a single line.

Output

Print exactly one line for each test case. The line should contain an integer representing the number of ways.

Sample Input

3
4
7
10

Sample Output

7
44
274

 

Problem Source: SJTU Programming Contest 2004

#include <stdio.h>
int sum;
void f(int n)
{
 if(n>=3)
  f(n-3);
 if(n>=2)
  f(n-2);
 if(n>=1)
  f(n-1);
 if(n == 0)
  sum++;
}
int main()
{
 int n,num;
 scanf("%d",&n);

 while(n>0)
 {
  sum = 0;
  scanf("%d",&num);
  f(num);
  printf("%d/n",sum);
  n--;
 }
return 0;  
}

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