R的練習

[jackdan (Sean)]

tmp<-function(m, n)
{
  cat("   方法1\n")
  ##m到n的總和 m到n的連乘 方法1##
  S=0
  P=1
  for(i in m:n)
  {
    S=S+i
    P=P*i
  }
  cat(m, "到", n, "的總和為", S, "\n")
  cat(m, "到", n, "的連乘為", P, "\n")

  cat("   方法2\n")
  ##m到n的總和 m到n的連乘 方法2##
  cat(m, "到", n, "的總和為", sum(m:n), "\n") #m:n 為(m, m+1, ..., n)的陣列
  cat(m, "到", n, "的連乘為", prod(m:n), "\n") #prod(m:n)=m*(m+1)*...*n

  cat("   方法1\n")
  ##m到n的偶數和與積數和##
  even=0
  odd=0
  for(i in m:n)
  {
    if(i%%2==0) even=even+i
    else  odd=odd+i
  }
  cat(m, "到", n, "的偶數和為", even, "\n")
  cat(m, "到", n, "的積數和為", odd, "\n")

  cat("   方法2\n")
  ##m到n的偶數和與積數和 方法2##
  if(m%%2==0)
  {
    cat(m, "到", n, "的偶數和為", sum(seq(m, n, 2)), "\n")
    cat(m, "到", n, "的積數和為", sum(seq(m+1, n, 2)), "\n")
  }
    #seq(a, b, c)為 a, a+c, a+2c, ..., a+n*c 其中a+n*c<=b 的陣列
    #例如 seq(1, 10, 2) 為 1, 3, 5, 7, 9
    #    seq(2, 10, 2) 為 2, 4, 6, 8, 10
  else
  {
    cat(m, "到", n, "的偶數和為", sum(seq(m+1, n, 2)), "\n")
    cat(m, "到", n, "的積數和為", sum(seq(m, n, 2)), "\n")
  }

  cat("\n")
  ##m到n中能被2或3整除的整數和  m到n中能被2與3整除的整數和##
  a=0
  b=0
  for(i in m:n)
  {
    if(i%%2==0 || i%%3==0) a=a+i
    if(i%%2==0 && i%%3==0) b=b+i
  }
  cat(m, "到", n, "中能被2或3整除的整數和為", a, "\n")
  cat(m, "到", n, "中能被2與3整除的整數和為", b, "\n")
}

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