题解 | #百钱买百鸡问题#

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

// 方法一：暴力法
void async function () {
    while(line = await readline()){
        for(let i = 0; i < 20; i++){
            for(let j = 0; j < 33; j++){
                for(let k = 0; k < 300; k++){
                    if(i*15+j*9+k*1 === 300 && i+j+k==100) console.log(i,j,k);
                }
            }
        }
    }
}()

// 方法二：数学
// i*15+j*9+k==300
// i+j+k==100
// i = (100-4*j)/7 或 j = (100-7*i)/4 = 25-7*i/4
// 25-7*i/4可知，i是4的倍数，且i的范围是0到100/5=20,所以i取0,4,8,12,16,20
void async function () {
    while(line = await readline()){
        for(let i = 0; i <=20; i+=4){
            const j = (100-7*i)/4;
            const k = 100 - i - j;
            if(j<0 || k<0) break;
            console.log(i,j,k);
        }        
    }
}()