问题在于您选择终止for循环的方式：break仅停止内部循环。请尝试以下代码，并检查p和q的值，执行将在该值处停止：

M=input('Give an upper bound for p:')
 N=input('Give an upper bound for q:')
 d=input('Give a positive integer between 1 and 15:')

 for q=1:N
 for p=1:M
   if abs(pi-p/q)<=10^(-d)
     [p,q]
     error('Found!') % Arrest the program when condition is met
 end
 end
 end

其给出以下输出：

当然，有更好的方法来捕获所有满足条件的整数对（例如，使用disp代替error），这超出了本答案的范围，但我将在这里提供几个例子：

clear; clc;

M=input('Give an upper bound for p:')
N=input('Give an upper bound for q:')
d=input('Give a positive integer between 1 and 15:')

pairs = []; % p,q pairs are stored here
k = 1; % counter

for q=1:N
    for p=1:M
        if abs(pi-p/q)<=10^(-d)
            pairs(k,1) = p; 
            pairs(k,2) = q;
            k = k + 1; % increment the counter
        end
    end
end

上面的脚本将安静地结束：（p，q）对将被存储在pair矩阵中。
下面的代码将直接打印这些对：

clear; clc;

M=input('Give an upper bound for p:')
N=input('Give an upper bound for q:')
d=input('Give a positive integer between 1 and 15:')

for q=1:N
    for p=1:M
        if abs(pi-p/q)<=10^(-d)
            out = sprintf("(%d, %d)", p,q);
            disp(out);
        end
    end
end

为了实验的需要，并根据@Cris Luengo的评论，这里有一个稍微更详细的脚本版本：x1M7 N1 x循环被封装在专用函数中，并且x1M8 N1 x循环很好地跟踪进程，并且用x1M10 N1 x对填充x1M9 N1 x矩阵：

clear; clc;

M=input('Give an upper bound for p:');
N=input('Give an upper bound for q:');
d=input('Give a positive integer between 1 and 15:');

close_to_pi = @(px,qx) abs(pi-px/qx)<=10^(-d); % returns true/false

p = 1; q = 1;
count = 0;

res = nan(N*M,2) ; % (p,q) pairs are stored here; preallocate for speed!

tic % starts the timer
while (q <= N)
    [p,q, found] = approx_pi(p, q, N, M, close_to_pi);
    
    if found
        count = count + 1;
        res(count,:) = [p,q]; % populates output var

    end
    
    if p<M % we aren't done with p
        p = p + 1;
    else % reset p and increment q
        p = 1;
        q = q + 1;
    end

    
end
res(isnan(res(:,1)),:) = []; % gets rid of the empty elements
disp(count)
toc % stops the timer and prints elapsed time

function [p, q, found] = approx_pi(p0, q0, N, M, fun)

for q=q0:N
    for p=p0:M
        if fun(p,q)
            found = 1;
            return
        end % if
    end % for p
    p0 = 1;
end % for q

found = 0;

end % approx_pi

如果您对pi的连续分数近似值感兴趣，请尝试rat(pi, Tol)，其中Tol是公差。