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48 lines
2.0 KiB
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48 lines
2.0 KiB
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<p>请你实现三个 API <code>append</code>，<code>addAll</code> 和 <code>multAll</code> 来实现奇妙序列。</p>
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<p>请实现 <code>Fancy</code> 类 ：</p>
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<ul>
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<li><code>Fancy()</code> 初始化一个空序列对象。</li>
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<li><code>void append(val)</code> 将整数 <code>val</code> 添加在序列末尾。</li>
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<li><code>void addAll(inc)</code> 将所有序列中的现有数值都增加 <code>inc</code> 。</li>
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<li><code>void multAll(m)</code> 将序列中的所有现有数值都乘以整数 <code>m</code> 。</li>
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<li><code>int getIndex(idx)</code> 得到下标为 <code>idx</code> 处的数值（下标从 0 开始），并将结果对 <code>10<sup>9</sup> + 7</code> 取余。如果下标大于等于序列的长度，请返回 <code>-1</code> 。</li>
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</ul>
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<p> </p>
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<p><strong>示例：</strong></p>
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<pre>
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<strong>输入：</strong>
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["Fancy", "append", "addAll", "append", "multAll", "getIndex", "addAll", "append", "multAll", "getIndex", "getIndex", "getIndex"]
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[[], [2], [3], [7], [2], [0], [3], [10], [2], [0], [1], [2]]
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<strong>输出：</strong>
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[null, null, null, null, null, 10, null, null, null, 26, 34, 20]
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<strong>解释：</strong>
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Fancy fancy = new Fancy();
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fancy.append(2); // 奇妙序列：[2]
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fancy.addAll(3); // 奇妙序列：[2+3] -> [5]
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fancy.append(7); // 奇妙序列：[5, 7]
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fancy.multAll(2); // 奇妙序列：[5*2, 7*2] -> [10, 14]
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fancy.getIndex(0); // 返回 10
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fancy.addAll(3); // 奇妙序列：[10+3, 14+3] -> [13, 17]
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fancy.append(10); // 奇妙序列：[13, 17, 10]
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fancy.multAll(2); // 奇妙序列：[13*2, 17*2, 10*2] -> [26, 34, 20]
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fancy.getIndex(0); // 返回 26
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fancy.getIndex(1); // 返回 34
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fancy.getIndex(2); // 返回 20
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 <= val, inc, m <= 100</code></li>
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<li><code>0 <= idx <= 10<sup>5</sup></code></li>
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<li>总共最多会有 <code>10<sup>5</sup></code> 次对 <code>append</code>，<code>addAll</code>，<code>multAll</code> 和 <code>getIndex</code> 的调用。</li>
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</ul>
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