c++比较简单的问题，求解！

次对角线和有问题
minor_sum+=a[m-1][i]应该改为minor_sum+=a[m-i-1][i]

你的max=min=a[0][0]比较好啊，矩阵也可以为负啊，minor_sum+=a[i][m-i-1]

楼上把bug都说完了。你可以在给max和min赋值的时候顺便记录下其坐标。这样就不需要第二次遍历了。

 int _tmain(int argc, _TCHAR* argv[])
{
    int m = 5, n = 5;
    int a[5][5] = {0};

    int i = 0, j = 0;
    for (i = 0; i < m; i++)
    {
        for (j = 0; j < n; j++)
        {
            cin >> a[i][j];
        }
    }

    for (i = 0; i < m; i++)
    {
        for (j = 0; j < n; j++)
        {
            cout << a[i][j] << " ";
        }
        cout <<endl;
    }

    int nMin = a[0][0], nMax = a[0][0];
    int nMain_Sum = 0, nMinor_Sum = 0;
    int nRowMin = 0, nColMin = 0;
    int nRowMax = 0, nColMax = 0;
    for (i = 0; i < m; i++)
    {
        for (j = 0; j < n; j++)
        {
            int nTem = a[i][j];
            if (nMin > nTem)
            {
                nMin = nTem;
                nRowMin = i;
                nColMin = j;
            }

            if (nMax < nTem)
            {
                nMax = nTem;
                nRowMax = i;
                nColMax = j;
            }

        }

        nMain_Sum += a[i][i];
        nMinor_Sum += a[m-i-1][i];
    }

    cout<< "最大值为:" << nMax << "在第" << nRowMax << "行" << "第" << nColMax << "列" << endl;
    cout<< "最小值为:" << nMin << "在第" << nRowMin << "行" << "第" << nColMin << "列" << endl;
    cout<< "主对角线之和为:" << nMain_Sum <<endl;
    cout<< "次对角线之和为:" << nMinor_Sum <<endl;