Write a function called one_per_n that returns the smallest positive integer n for which the sum 1 + 1/2 + 1/3 + … + 1/n , is greater than or equal to x where x is the input argument. Limit the maximum number n of terms in the sum to 10,000 .

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prasanta sahoo
prasanta sahoo 2015-5-22
回答： Jorge Briceño 2018-1-29
function out=one_per_n(x)
if x>10000
out=-1;
%fprintf('-1\n');
else
total=0;
for n=1:x;
total=(total+(1/n));
end
if total>=x
out=total;
%fprintf('%d\n',total);
end
end

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采纳的回答

Walter Roberson
Walter Roberson 2015-5-22
Example:
for K = 1 : 50 %maximum 50 times
t = f(K); %some calculation dependent on the counter
if t >= SomeLimit %did we reach the limit?
break; %Yes! Stop this loop!
end
end
%now that we got here you need to figure out if you matched what you wanted to match, or if you ran out of iterations
%left as an exercise for the reader
  4 个评论
显示 2更早的评论
Christos Vyzantios
Christos Vyzantios 2015-5-31
编辑：Walter Roberson 2015-6-1
The solution seems ok but the greater saw us the message:
Problem 5 (one_per_n):
Feedback: Your function performed correctly for argument(s) 1
Feedback: Your function performed correctly for argument(s) 2
Feedback: Your function performed correctly for argument(s) 3
Feedback: Your function performed correctly for argument(s) 4
Feedback: Your function performed correctly for argument(s) 8
Feedback: Your function performed correctly for argument(s) 9
Feedback: Your function performed correctly for argument(s) 9.7875
Feedback: Your function performed correctly for argument(s) 9.7876
Feedback: Your function made an error for argument(s) 9.7877
Your solution is _not_ correct.
Walter Roberson
Walter Roberson 2015-6-1
The problem statement from the title does not define what should happen when 10000 terms is not enough to reach the desired total.

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更多回答（4 个）

Chen Li
Chen Li 2017-3-17
编辑：Chen Li 2017-3-17
function d = holiday(month,day)
if month == 1 && day == 1
d = true;
elseif month == 7 && day == 4
d = true;
elseif month == 12 && day == 25
d = true;
elseif month == 12 && day == 31
d = true;
else
d = false;
end
This is the code that is based on what you wrote. Now it works!
Jean Araujo
Jean Araujo 2015-6-1
function out = one_per_n(x)
total=0;
n = 1;
while total < x
total = (total+1/n);
n = n+1;
end
if n-1 > 10000
out = -1;
else
out = (n-1);
end
Nava Subedi
Nava Subedi 2016-11-28
编辑：Walter Roberson 2016-11-28
I tried to solve the question.
Write a function called holiday that takes two input arguments called month and day; both are scalar integers representing a month (1-12) and a day (1-31). The function returns a logical true if the specified date is a holiday; if not, it returns false. For the purposes of this exercise, the following dates are considered holidays: January 1st, July 4th, December 25th, and December 31st.
I wrote:
function d = holiday(month, day)
if (month >= 1 && month <=12) && (day >= 1 && day <= 31)
if month == 1 && day == 1
d = 1
fprintf('%d is true! It is a holoday. \n', d);
end
if month == 7 && day == 4
d = 1
fprintf('%d is true! It is a holoday. \n', d);
end
if month == 12 && day == 25
d = 1
fprintf('%d is true! It is a holoday. \n', d);
end
if month == 12 && day == 31
d = 1
fprintf('%d is true! It is a holoday. \n', d);
end
else
d = 0
fprintf('%d is false! It is not a holoday. \n', d);
return
end
end
But did not work. Where did I do mistake??
  1 个评论
Walter Roberson
Walter Roberson 2016-11-28
The assignment does not request any printing.
The assignment requires logical true and logical false as outputs, not double precision 1 and double precision 0. Logical true is represented as true and logical false is represented as false

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Jorge Briceño
Jorge Briceño 2018-1-29
Hello Nava,
Maybe you could also try to simplify your code. Here is my example:
function out = holiday (m, d)
if (m==1 && d==1) || (m==7 && d==4)|| (m==12 && d==25)...
|| (m==12 && d==31)
out=true;
else
out=false;
end
return;
end

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