[SPOJ] ABCDEF – ABCDEF | 书脊

[SPOJ] ABCDEF – ABCDEF | 书脊
http://www.spoj.com/problems/ABCDEF/

You are given a set S of integers between -30000 and 30000 (inclusive).

Find the total number of sextuples  that satisfy: 

Input

The first line contains integer N (1 ≤ N ≤ 100), the size of a set S.

Elements of S are given in the next N lines, one integer per line. Given numbers will be distinct.

Output

Output the total number of plausible sextuples.

Examples

Input: 1 1 Output: 1

Input: 2 2 3 Output: 4

Input: 2 -1 1 Output: 24

Input: 3 5 7 10 Output: 10

题目大意:计算一个集合S,范围在[-30000,30000], 其中所有的可能的整数abcdef. 满足: (a*b+c)/d-e=f, where d !=0. 问多少种情况

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分析: 公式题, 首先是整数取值, 所以就不难.先化简成a*b+c=d(e+f), 然后数一下左边集合中每个在右边的的个数. 注意:

1. 这里是每个元素, 包括重复的.
2. SPOJ真是卡空间, 用map直接TLE, 好好自己写counter吧.
3. d不能是0

http://neverendingbits.blogspot.com/2011/12/spoj-4580-abcdef.html

The Proper Approach:

N<=100 => N^3<=10^6 numbers which can easily be stored in an array.

Rearrange the equation as (a*b+c)=d*(e+f)

• Find the values of (a*b+c) for all possible sets (a,b,c) and store them in an array arr1 --> O(N^3) time.
• Sort arr1 --> O(N^3 log(N^3)) = O(N^3 logN) time.
• Find the values of (d*(e+f)) for all possible sets (d,e,f) and store them in an array arr2 --> O(N^3) time.
• Sort arr2 --> O(N^3 log(N^3)) = O(N^3 logN) time.
• Now perform an intersection operation on the two arrays, arr1 and arr2, realizing that two same values in an array come from two distinct subsets of either (a,b,c) or (d,e,f) and thus contribute to two different sets (a,b,c,d,e,f).

This all takes O(N^3 logN) time (about 10^7 operations) manageable within a second :)

public void solve(int testNumber, InputReader in, OutputWriter out) {

        int n = in.readInt();

        int[] nums = new int[n];

        for (int i = 0; i < n; i++) {

            nums[i] = in.readInt();

        }

        Arrays.sort(nums);

        int[] left = new int[n*n*n];

        int p = 0;

        for (int i = 0; i < nums.length; i++) {

            for (int j = 0; j < nums.length; j++) {

                for (int k = 0; k < nums.length; k++) {

                    left[p++] = nums[i] * nums[j] + nums[k];

                }

            }

        }

        Arrays.sort(left, 0, n*n*n);

        int[][] range = new int[2][left.length];

        p = 0;

        long count = 0;

        int tmp = Integer.MAX_VALUE;

        for (int i = 0; i < left.length; i++) {

            if (tmp != left[i]){

                range[0][p] = left[i];

                range[1][p]++;

                p++;

            }

            else {

                range[1][p-1]++;

            }

            tmp = left[i];

        }

        for (int i = 0; i < nums.length; i++) {

            for (int j = 0; j < nums.length; j++) {

                for (int k = 0; k < nums.length; k++) {

                    if (nums[i] == 0)

                        continue;

                    else {

                        int right = nums[i]*(nums[j]+nums[k]);

                        int b = Arrays.binarySearch(range[0], 0, p, right);

                        if (b >= 0)

                            count+= range[1][b];

                    }

                }

            }

        }

        out.print(count);

    }

https://github.com/srikk595/Spoj-Problems/blob/master/ABCDEF-6726594-src.cpp
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n;i++)
    {
        scanf("%d",&x[i]);
    }
    for (int i=0;i<n;i++)
    {
        for (int j=0;j<n;j++)
        {
            for (int k=0;k<n;k++)
            {
                s1.push_back(x[i]*x[j]+x[k]);
            }
        }
    }
    for (int i=0;i<n;i++)
    {
        for (int j=0;j<n;j++)
        {
            for (int k=0;k<n;k++)
            {
                if (x[k]==0) continue;
                s2.push_back((x[i]+x[j])*x[k]);
            }
        }
    }
    sort(s1.begin(),s1.end());
    sort(s2.begin(),s2.end());
    for (int i=0;i<s1.size();i++)
    {
        lo=lower_bound(s2.begin(),s2.end(),s1[i])-s2.begin();
        hi=upper_bound(s2.begin(),s2.end(),s1[i])-s2.begin();
        res+=(hi-lo);
    }
    printf("%lld\n",res);
    return 0;
}
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