ODZT：判断一组不等式是否满足约束并输出最大差 - A

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描述

给定一组不等式，判断是否成立并输出不等式的最大差(输出浮点数的整数部分)，要求：1）不等式系数为double类型，是一个二维数组；2）不等式的变量为int类型，是一维数组；3）不等式的目标值为double类型，是一维数组；4）不等式约束为字符串数组，只能是：">",">=","<","<=","="，例如,不等式组：
a11*x1+a12*x2+a13*x3+a14*x4+a15*x5<=b1;
a21*x1+a22*x2+a23*x3+a24*x4+a25*x5<=b2;

a31*x1+a32*x2+a33*x3+a34*x4+a35*x5<=b3;

最大差=max{ (a11*x1+a12*x2+a13*x3+a14*x4+a15*x5-b1), (a21*x1+a22*x2+a23*x3+a24*x4+a25*x5-b2), (a31*x1+a32*x2+a33*x3+a34*x4+a35*x5-b3) }，类型为整数(输出浮点数的整数部分)

输入描述：

1、不等式组系数(double类型)：

a11,a12,a13,a14,a15
a21,a22,a23,a24,a25
a31,a32,a33,a34,a35

2、不等式变量(int类型)：

x1,x2,x3,x4,x5

3、不等式目标值(double类型)：b1,b2,b3

4、不等式约束(字符串类型):<=,<=,<=
a11,a12,a13,a14,a15;a21,a22,a23,a24,a25;a31,a32,a33,a34,a35;x1,x2,x3,x4,x5;b1,b2,b3;<=,<=,<=

输出描述：

示例 1

输入

2.3,3,5.6,7,6;11,3,8.6,25,1;0.3,9,5.3,66,7.8;1,3,2,7,5;340,670,80.6;<=,<=,<=

输出

false 458

示例 2

输入

2.36,3,6,7.1,6;1,30,8.6,2.5,21;0.3,69,5.3,6.6,7.8;1,13,2,17,5;340,67,300.6;<=,>=,<=

输出

false 758

解题思路

1、对输入数据用”;”进行分割，分割后的的数组最后三个数组分别为不等式变量、不等式目标值，不等式约束，剩下的都是不等式系数
2、然后根据题意对系数和变量进行×、+处理
3、通过目标值和约束进行判断处理，并获取最大值

JavaScript

let strings = readLine().split(";");
//let strings = "2.36,3,6,7.1,6;1,30,8.6,2.5,21;0.3,69,5.3,6.6,7.8;1,13,2,17,5;340,67,300.6;<=,>=,<=".split(";");

let length = strings.length;
let bianliang = strings[length - 3].split(",").map((i) => Number(i)); //不等式变量
let mubiao = strings[length - 2].split(",").map((i) => Number(i)); //不等式目标值
let ys = strings[length - 1].split(","); //不等式约束

let m = ys.length; //约束的数量等于数组的数量
let n = bianliang.length; //变量的数量等于数组中数据的数量

let doubles = []; //不等式系数是二维数组

for (let i = 0; i < m; i++) {
    let xs = strings[i].split(","); //不等式系数
    doubles[i] = new Array();
    for (let j = 0; j < n; j++) {
        doubles[i][j] = Number(xs[j]); //将不等式系数放入double类型二维数组
    }
}

let isYueshu = true;
let max = 0;

for (let i = 0; i < m; i++) {
    //循环遍历不等式数组
    let d = 0;
    let b = true;
    for (let j = 0; j < n; j++) {
        d += doubles[i][j] * bianliang[j]; //不等式数组值
    }
    max = Math.max(max, d - mubiao[i]); //求出最大差
    if (ys[i] == ">") {
        //等于不等式进行判断
        b = d > mubiao[i];
    } else if (ys[i] == ">=") {
        b = d >= mubiao[i];
    } else if (ys[i] == "<") {
        b = d < mubiao[i];
    } else if (ys[i] == "<=") {
        b = d <= mubiao[i];
    } else if (ys[i] == "=") {
        b = d == mubiao[i];
    }
    if (!b) {
        //只要一个等式不成立就为false
        isYueshu = false;
    }
}

console.log(isYueshu + " " + parseInt(max));