简单递归

放苹果

把M个同样的苹果放在N个同样的盘子里，允许有的盘子空着不放，问共有多少种不同的分法？（用K表示）5，1，1和1，5，1 是同一种分法。Input

第一行是测试数据的数目t（0 <= t <= 20）。以下每行均包含二个整数M和N，以空格分开。1<=M，N<=10。

Output

对输入的每组数据M和N，用一行输出相应的K。

Sample Input

1
7 3

Sample Output

8

#include<iostream>
#include<cstring>
using namespace std;
int a[21][21];
int apple(int m,int n)
{
    if(a[m][n])return a[m][n];
    else if(m==1 || n==1)
    {
        a[m][n]=1;
        return a[m][n];
    }
    else if(m<n)
    {
        a[m][n]=apple(m,m);
        return a[m][n];
    }
    else if(m==n)
    {
        a[m][n]=1+apple(m,m-1);
        return a[m][n];
    }
    else
    {
        a[m][n]=apple(m-n,n)+apple(m,n-1);
        return a[m][n];
    }
} 
int main()
{
    int T,m,n;
    cin>>T;
    while(T--)
    {
        memset(a,0,sizeof(a));
        cin>>m>>n;
        cout<<apple(m,n)<<endl;
    }
    return 0;
}

母牛的故事

有一头母牛，它每年年初生一头小母牛。每头小母牛从第四个年头开始，每年年初也生一头小母牛。请编程实现在第n年的时候，共有多少头母牛？Input输入数据由多个测试实例组成，每个测试实例占一行，包括一个整数n(0<n<55)，n的含义如题目中描述。
n=0表示输入数据的结束，不做处理。Output对于每个测试实例，输出在第n年的时候母牛的数量。
每个输出占一行。Sample Input

2
4
5
0

Sample Output

2
4
6

#include<iostream>
using namespace std;
long long  a[100];
int main()
{
  int n;
  while(cin>>n)
  {
    a[1]=1;a[2]=2;a[3]=3;a[4]=4;
    if(n==0) break;
    if(n<=4) cout<<n<<endl;
    else
    {
      for(int i=5;i<=n;i++)
      {
        a[i]=a[i-1]+a[i-3];
      }
      cout<<a[n]<<endl;
    }
  }
  return 0;
}

Fibonacci Again                    

There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).
InputInput consists of a sequence of lines, each containing an integer n. (n < 1,000,000).
OutputPrint the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.
Sample Input

0
1
2
3
4
5

Sample Output

no
no
yes
no
no
no

#include<iostream>
using namespace std;
int a[10000004];
int main()
{
  int n;
  a[0]=1;
  a[1]=2;
  for(int i=2;i<=1000000;i++)
  {
    a[i]=(a[i-1]+a[i-2])%3;
  }
  while(cin>>n)
  {
    if(a[n]==0) cout<<"yes"<<endl;
    else cout<<"no"<<endl;
  }
  return 0;
}

Children’s Queue                    

There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?
InputThere are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)OutputFor each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.Sample Input

1
2
3

Sample Output

1
2
4

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=1100;
int a[maxn][maxn];
int main()
{
  int n,i,j;
  memset(a,0,sizeof(a));
  a[0][0]=0;
  a[1][0]=1;
  a[2][0]=2;
  a[3][0]=4;
  a[4][0]=7;
  for(i=5;i<=1001;i++)
  {
    for(j=0;j<=1007;j++)
    {
      a[i][j]+=(a[i-1][j]+a[i-2][j]+a[i-4][j]);
      if(a[i][j]>=10)
      {
        int h=a[i][j];
        a[i][j]=h%10;
        a[i][j+1]+=(h/10);
      }
    }
  }
  while(cin>>n)
  {
    for(i=1007;i>=0;i--)
    {
      if(a[n][i]) break;
    }
    for( ;i>=0;i--)
    {
      cout<<a[n][i];
    }
    cout<<endl;
  }
  return 0;

}

一只小蜜蜂...                    

有一只经过训练的蜜蜂只能爬向右侧相邻的蜂房，不能反向爬行。请编程计算蜜蜂从蜂房a爬到蜂房b的可能路线数。
其中，蜂房的结构如下所示。

Input输入数据的第一行是一个整数N,表示测试实例的个数，然后是N 行数据，每行包含两个整数a和b(0<a<b<50)。
Output对于每个测试实例，请输出蜜蜂从蜂房a爬到蜂房b的可能路线数，每个实例的输出占一行。
Sample Input

2
1 2
3 6

Sample Output

1
3

#include<iostream>
using namespace std;
__int64 a[52];
int main()
{
  __int64 N,c,b,i;
  a[0]=1;
  a[1]=1;
  for(i=2;i<=50;i++)
  {
    a[i]=a[i-1]+a[i-2];
  }
  cin>>N;
  while(N--)
  {
    cin>>c>>b;
    cout<<a[b-c]<<endl;
  }
  return 0;
}