Divide and conquer:4 Values whose Sum is 0(POJ 2785)

　　　　　　　　　　　　　　　　　

　　　　　　　　　　　　　　　　找四个数的和为0

　　题目大意：给定四个集合，要你每个集合选4个数字，组成和为0

　　这题是3977的简单版，只要和是0就可以了

　　

#include <iostream>
#include <algorithm>
#include <functional>
#define MAX 4001

using namespace std;

typedef long long LL_INT;
static LL_INT list[4][MAX], set_sum1[MAX*MAX];

LL_INT *Binary_Lower_Bound(int &, LL_INT);
LL_INT *Binary_Upper_Bound(int &, LL_INT);
LL_INT solve(int &,int &);

int main(void)
{
    int list_contian, sum_comb;
    LL_INT ans;

    while (~scanf("%d", &list_contian))
    {
        for (int i = 0; i < list_contian; i++)
            scanf("%lld%lld%lld%lld", &list[0][i], &list[1][i], &list[2][i], &list[3][i]);

        sum_comb = 0;
        for (int i = 0; i < list_contian; i++)//把第一张和第二张表的总数枚举出来
            for (int j = 0; j < list_contian; j++)
                set_sum1[sum_comb++] = list[0][i] + list[1][j];

        sort(set_sum1, set_sum1 + sum_comb);
        ans = solve(list_contian, sum_comb);
        printf("%lld
", ans);
    }
    return EXIT_SUCCESS;
}

LL_INT solve(int &list_contain, int &sum_comb)
{
    LL_INT tmp_sum, ans = 0;
    int pos1, pos2;

    for (int i = 0; i < list_contain; i++)
    {
        for (int j = 0; j < list_contain; j++)
        {
            tmp_sum = -list[2][i] - list[3][j];
            pos1 = Binary_Upper_Bound(sum_comb, tmp_sum) - set_sum1;
            pos2 = Binary_Lower_Bound(sum_comb, tmp_sum) - set_sum1;
            ans += pos1 - pos2;
        }
    }
    return ans;
}

LL_INT *Binary_Lower_Bound(int &sum_comb, LL_INT val)
{
    int lb = 0, mid, count1 = sum_comb, count2;
    while (count1 > 0)
    {
        count2 = count1 >> 1;
        mid = lb + (count1 >> 1);
        if (set_sum1[mid] < val)
        {
            lb = ++mid;
            count1 -= count2 + 1;
        }
        else count1 = count2;
    }
    return &set_sum1[lb];
}

LL_INT *Binary_Upper_Bound(int &sum_comb, LL_INT val)
{
    int lb = 0, mid, count1 = sum_comb, count2;
    while (count1 > 0)
    {
        count2 = count1 >> 1;
        mid = lb + (count1 >> 1);
        if (set_sum1[mid] <= val)
        {
            lb = ++mid;
            count1 -= count2 + 1;
        }
        else count1 = count2;
    }
    return &set_sum1[lb];
}