题解 | #称砝码#普通循环遍历和回溯法，回溯法时间复杂度大

const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;

void async function () {
    // Write your code here
    while(n = Number(await readline())){
        let weights = (await readline()).split(' ').map(Number);
        let nums = (await readline()).split(' ').map(Number);
        // 把所有砝码放在一个数组，遍历这个数组，在改遍历中再遍历res的每一项，然后加weights[i]
        weights = weights.reduce((pre, cur, i) => pre.concat(new Array(nums[i]).fill(cur)), []);
        let res = new Set([0]);
        for (let i = 0; i < weights.length; i++) {
            [...res].forEach(v => {
                res.add(v + weights[i]);
            });
        }
        // 这里用回溯法，时间复杂度大，数据多的时候时间过不了，看看哪个大神有优化方案
        // let used = new Set();
        // backtrack(0, 0);
        // function backtrack(index, sum) {
        //     res.add(sum);
        //     if (index >= weights.length) return;
        //     for (let i = index; i < weights.length; i++) {
        //         if (!used.has(i)) {
        //             used.add(i);
        //             backtrack(i, sum + weights[i])
        //             used.delete(i);
        //         }
        //     }
        // }
        console.log(res.size);
    }
}()