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    <p>Since 100 / 50 is an integer, this is quite easy. Since 50 / (100 / 50) is an integer, it's even easier.</p> <p>If you didn't quite get that, here is some sample code:</p> <pre><code>int rnd1 = getrnd50(); int rnd2 = getrnd50(); if (rnd1 % 2 == 0) { rnd2 += 50; } return rnd2; </code></pre> <p>Here is an outline:</p> <ul> <li>Two numbers, chosen randomly between 1 and 50, called <strong>a</strong> and <strong>b</strong>.</li> <li>If <strong>a</strong> is even, add 50 to <strong>b</strong>.</li> <li>Return <strong>b</strong>.</li> </ul> <p>You can make this a one-liner if you want:</p> <pre><code>return getrnd50() + getrnd50() % 2 * 50; </code></pre> <p>That's a little too obfuscated though.</p> <p>Edit: I see the question was really asking for a shuffled list, not a sequence of random integers.</p> <p>This can be done by creating a list from 1 to 100, and doing 100 random swaps, like a Fisher-Yates shuffle. I imagine that with a Fisher-Yates shuffle, the absolute minimum number of calls is 93 (given with the formula <code>ceil(log50(100!))</code>), but with a much simpler algorithm you can use 200.</p> <p>The simple algorithm would involve swapping each of the 100 elements with a random element from the 100. The number to choose would be generated from 1-100 with the above generator.</p> <p>For example:</p> <pre><code>for (int i = 0; i &lt; 100; i++) { swap(i, getrnd100() - 1); // - 1 for zero base! } </code></pre> <p>Here is some complete code:</p> <pre><code>int[] result = new int[100]; for (int i = 0; i &lt; 100; i++) { result[i] = i + 1; } for (int i = 0; i &lt; 100; i++) { int j = (getrnd50() + getrnd50() % 2 * 50) - 1; int tmp = result[i]; result[i] = result[j]; result[j] = tmp; } return result; </code></pre> <p>(Disclaimer: I don't know Java, and I haven't tested it.)</p> <p>Best case 200, worst case 200, average case 200.</p>
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    1. POGenerate random numbers from 1-100 from a generator of 1-50
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    1. COWastes two generations for each output. OP got criticized for that.
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    2. COthis does not ensure, that every number from 1 -100 hundred is printed in a finite amount of calls
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    3. CO@MarkoTopolnik Wastes two? It only uses two.
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