Question

我遇到了一个奇怪的问题。添加两个绝对（正）值时，我得到一个负值。我试图通过编码来解决这个练习（链接：https://www.codingame.com/ide/puzzle/network-cabling）。

我试过调试，我得到了这个：

X[i]: 19715507  X[j]: 938059973   // This one is good
temp: 918344466   

Y[i]: 470868309  Y[j]: -816049599 //Something is wrong here
temp: -2089704922

这段代码可能不是一个很好的解决方案，我需要改进它，但我仍然无法弄清楚为什么价值是负面的。

请帮忙。

谢谢。

代码：

int main()
{

    int N;
    int X[100000];
    int Y[100000];

    cin >> N; cin.ignore();
    for (int i = 0; i < N; i++) {   //reading all input

        cin >> X[i] >> Y[i]; cin.ignore();
    }

    int ilgis=0;   //ilgis means length

    for(int i=0; i<N-1; i++){       //"min" is set to a very high value 
                                    //so it could find a lower value later
                                    // "not the best solution"

        int min=99999999999999999999;   //shortest distance between houses
        for(int j=0; j<N; j++){    // here I am trying to find the shortest 
                                   // distance from one house to another

            if(j!=i){     //I can't count distance from the same house because 
                          // it would be 0

                int temp=0;
                                    //counting the distance differently if the 
                                    //value is negative or positive

                if(X[i]<=0&&X[j]<=0) temp+=abs(abs(X[i])-abs(X[j]));
                else if(X[i]<=0&&X[j]>=0) temp+=abs(X[i])+abs(X[j]);
                else if(X[i]>=0&&X[j]>=0) temp+=abs(X[i]-X[j]);
                else if(X[i]>=0&&X[j]<=0) temp+=(X[i])+abs(X[j]);

                                       //same with y axis

                if(Y[i]<=0&&Y[j]<=0) temp+=abs(abs(Y[i])-abs(Y[j]));
                else if(Y[i]<=0&&Y[j]>=0) temp+=abs(Y[i])+abs(Y[j]);
                else if(Y[i]>=0&&Y[j]>=0) temp+=abs(Y[i]-Y[j]);
                else if(Y[i]>=0&&Y[j]<=0) temp+=(Y[i])+abs(Y[j]); 

                if(min>temp) min=temp;  
                }
            }                   //if i found the shortesst distance between 
                                //houses I add that value to overall distance
                                // and continue until all houses are checked
        ilgis+=min;

        }
    cout<<ilgis<<endl;
}

练习的输入：

 8
-28189131 593661218
102460950 1038903636
938059973 -816049599
-334087877 -290840615
842560881 -116496866
-416604701 690825290
19715507 470868309
846505116 -694479954

Answer 1

int的最大值为2147483647（假设int具有32位宽度，则为2 ^ 31-1）。您可以使用

进行检查

std::numeric_limits<int>::max();

您可能希望min使用此值。此外，还可以使用long long作为计算类型。

c ++在添加两个绝对（正）值时获得负值

1 个答案: