90 lines
2.2 KiB
Markdown
90 lines
2.2 KiB
Markdown
---
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id: 5900f3831000cf542c50fe96
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title: 问题23：非丰富的总和
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challengeType: 5
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videoUrl: ''
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dashedName: problem-23-non-abundant-sums
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---
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# --description--
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完美数字是一个数字，其正确除数的总和恰好等于数字。例如，28的适当除数之和为1 + 2 + 4 + 7 + 14 = 28，这意味着28是一个完美数。如果`n`的适当除数之和小于`n` ，则`n`被称为不足，如果该和超过`n`则称为`n` 。由于12是最小的有限数，1 + 2 + 3 + 4 + 6 = 16，可以写成两个有限数之和的最小数是24.通过数学分析，可以看出所有整数都大于28123可以写成两个数字的总和。然而，即使已知不能表示为两个充裕数的总和的最大数小于该限制，也不能通过分析进一步减小该上限。找出所有正整数<= `n`的总和，它不能写成两个丰富数字的总和。
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# --hints--
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`sumOfNonAbundantNumbers(10000)`应返回3731004。
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```js
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assert(sumOfNonAbundantNumbers(10000) === 3731004);
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```
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`sumOfNonAbundantNumbers(15000)`应该返回4039939。
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```js
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assert(sumOfNonAbundantNumbers(15000) === 4039939);
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```
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`sumOfNonAbundantNumbers(20000)`应返回4159710。
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```js
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assert(sumOfNonAbundantNumbers(20000) === 4159710);
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```
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`sumOfNonAbundantNumbers(28123)`应该返回4179871。
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```js
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assert(sumOfNonAbundantNumbers(28123) === 4179871);
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```
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# --seed--
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## --seed-contents--
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```js
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function sumOfNonAbundantNumbers(n) {
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return n;
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}
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sumOfNonAbundantNumbers(28123);
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```
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# --solutions--
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```js
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function abundantCheck(number) {
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let sum = 1;
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for (let i = 2; i <= Math.sqrt(number); i += 1) {
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if(number % i === 0) {
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sum += i + +(i !== Math.sqrt(number) && number / i);
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}
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}
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return sum > number;
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}
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function sumOfNonAbundantNumbers(n) {
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let sum = 0;
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const memo = {};
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let abundantList = [];
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// Function checkSum checks if num can be represented as a sum of numbers in the stack (array)
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const checkSum = (num, stack, memo) => {
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for (let i = 0; i < stack.length; i += 1) {
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if ((num - stack[i]) in memo) return true;
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}
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return false;
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};
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for (let i = 1; i <= n; i += 1) {
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if (abundantCheck(i)) {
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abundantList.push(i);
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memo[i] = 1;
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}
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if (checkSum(i, abundantList, memo)) continue;
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sum += i;
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}
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return sum;
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}
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```
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