题解 | #称砝码#
称砝码
https://www.nowcoder.com/practice/f9a4c19050fc477e9e27eb75f3bfd49c
动态规划解法 完整代码如下：
const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;
let lines = [];
void async function () {
// Write your code here
while(line = await readline()){
lines.push(line);
let n = parseInt(lines[0]);
if (lines.length == 3) {
let dp = Array(n + 1).fill([0]);
let arr1 = lines[1].split(' ').map(Number);
let arr2 = lines[2].split(' ').map(Number);
for (let i = 1; i <= n; i++) {
let index = 1;
let m = arr1[i-1];
let k = arr2[i-1];
dp[i] = dp[i].concat(Array(k).fill(0).map(x => m*(index++)));
}//以上的dp是将每个重量砝码单独的组合形式列了出来
for (let i = 1; i <= n; i++) {
dp[i] = expand(dp[i], dp[i-1]);//利用双重循环 对dp[i]重新赋值 更新之后的值为原dp[i]和dp[i-1]分别交叉相加之后的新数组，再利用set去重，则得到最终的dp[i]
}
console.log(dp[n].length)
}
}
}()
function expand(arr1, arr2) {
let res = [];
for (let i = 0; i < arr1.length; i++) {
for (let j = 0; j < arr2.length; j++) {
res.push(arr1[i] + arr2[j])
}
}
res = [...new Set(res)];
return res;
}
