如何在C或C ++或java中解决这个问题，请尽快回答这个问题 [英] How to do this problem in C or C++ or java, please answer this question as soon as possible

问题描述

任务是找到两个相同大小的数组的最小产品总和，因为在第一个数组上允许k个修改。在每次修改中，第一个数组的一个数组元素可以增加或减少2.



注意 - 乘积和是求和（A [i] * B [i]）对于从1到n的所有i，其中n是两个数组的大小

输入格式：

输入的第一行包含由空格分隔的n和k

第二行包含数组A（可修改的数组），其值由空格分隔

第三行包含数组B（不可修改的数组），其值由空格分隔



输出格式：



输出两个阵列的最小产品总和

约束条件：

1≤N≤10^ 5

0≤| A [i] |，| B [i] | ≤10^ 5

0≤K≤10^ 9







样本输入和输出





SNo。输入输出

1.



3 5

1 2 -3

-2 3 -5

-31



2.



5 3

2 3 4 5 4

3 4 2 3 2

25



我尝试过：

The task is to find the minimum sum of Products of two arrays of the same size, given that k modifications are allowed on the first array. In each modification, one array element of the first array can either be increased or decreased by 2.

Note- the product sum is Summation (A[i]*B[i]) for all i from 1 to n where n is the size of both arrays
Input Format:
First line of the input contains n and k delimited by whitespace
Second line contains the Array A (modifiable array) with its values delimited by spaces
Third line contains the Array B (non-modifiable array) with its values delimited by spaces

Output Format:

Output the minimum sum of products of the two arrays
Constraints:
1 ≤ N ≤ 10^5
0 ≤ |A[i]|, |B[i]| ≤ 10^5
0 ≤ K ≤ 10^9



Sample Input and Output


SNo. Input Output
1.

3 5
1 2 -3
-2 3 -5
-31

2.

5 3
2 3 4 5 4
3 4 2 3 2
25

What I have tried:

#include<iostream>
#include<math.h>
using namespace std;
int main()
{
    long long  int A[10^5],B[10^5];
    int N;
    int pro,temp,diff;
    int maxDiff=0,minSum=0;
    int K;
    cin>>N>>K;
    for(int i=0;i<n;i++)>
    {
        cin>>A[i];
        cin>>B[i];
    }
    for(int i=0;i<n;i++)>
    {
        pro=A[i]*B[i];
        if(pro<0 && B[i]<0)
        temp=(A[i]+2 * K)* B[i];
        else if(pro<0 && A[i]<0)
        temp=(A[i]-2 * K)* B[i];
        else if(pro>0 && A[i]<0)
        temp=(A[i]+ 2* K)* B[i];
        else if(pro>0 && A[i]>0)
        temp=(A[i]- 2*K) *B[i];
        diff=abs(pro-temp);
        if(diff>maxDiff)
        maxDiff=diff;
        minSum=minSum+pro;
    }

    minSum=minSum-maxDiff;
    cout<<"\n"<<minSum;

}

推荐答案

long long  int A[10^5],B[10^5];

以上甚至不会编译，如果确实如此，你将有两个数组，每个元素都是零。我建议阅读 C ++内置运算符，优先级和相关性 [ ^ ]。您还声明了一个变量 N ，但是然后尝试在中使用名为 n 的变量for 循环。对变量使用适当有意义的名称有助于避免诸如此类的简单错误。

The above will not even compile, and if it did you would have two arrays each of zero elements. I suggest a reading of C++ Built-in Operators, Precedence and Associativity[^]. You also declare a variable N, but then try to use a variable named n in your for loop. Using proper meaningful names for variables helps to avoid simple errors such as this.

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