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  • HDU 1081:To The Max

    To The Max

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 10747    Accepted Submission(s): 5149


    Problem Description
    Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

    As an example, the maximal sub-rectangle of the array:

    0 -2 -7 0
    9 2 -6 2
    -4 1 -4 1
    -1 8 0 -2

    is in the lower left corner:

    9 2
    -4 1
    -1 8

    and has a sum of 15.
     

    Input
    The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].
     

    Output
    Output the sum of the maximal sub-rectangle.
     

    Sample Input
    4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2
     

    Sample Output
    15
     

    Source
     

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    迷失在幽谷中的鸟儿,独自飞翔在这偌大的天地间,却不知自己该飞往何方……

    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    using namespace std;
    int fun(int b[110],int n)
    {
        int i,sum=0,max=-130;
        for (i = 1; i <= n; i++)
        {
            if (sum > 0)sum += b[i];
            else sum = b[i];
            if (max < sum)max = sum;
        }
        return max;
    }
    int main()
    {
        int i,j,k,sum,max,n,a[110][110], b[110];
        while (cin>>n)
        {
            sum = 0,max = -130;
            for (i = 1; i <= n; i++)
                for (j = 1; j <= n; j++)
                    scanf("%d", &a[i][j]);
            for (i = 1; i <= n; i++)
            {
                memset(b,0,sizeof(b));
                for (j = i; j <= n; j++)
                {
                    for (k = 1; k <= n; k++)
                        b[k]+=a[j][k];
                    sum = fun(b,n);
                    if (max < sum)max = sum;
                }
            }
            printf("%d
    ", max);
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/im0qianqian/p/5989408.html
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