zzuli 1785: 学渣的逆袭

http://acm.zzuli.edu.cn/problem.php?id=1785

学渣的逆袭

Description

 老师在黑板上写了四个数列a,b,c,d，数列a,b,c,d分别有i,j,k,l个数，突然间老师很生气的把正在睡觉的豆子喊了起来，问：“这是你第x次上课睡觉了！现在给你个赎罪的机会，你从每个数列中选择一个数，有多少种选法使他们的和为x？”，豆子实在太慌乱了，小伙伴们能告诉豆子正确答案吗？

Input

 第一行有四个整数i,j,k,l(1<=i,j,k,l<=500)，第二行有i个数a1,a2...ai，第三行有j个数b1,b2...bj，第四行有k个数c1,c2...ck，第五行有l个数d1,d2...dl。第六行有一个数m,接下来m行询问，每行有一个数字x。(m<=10),(|x|<10^10),(|a|,|b|,|c|,|d|<=10^8)

Output

 输出有m行，每行输出一个x对应的答案。

Sample Input

2 2 2 2
1 2
3 4
5 6
7 8
3
7
16
17

Sample Output

0
1
4

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;
typedef long long LL;

#define N 251100
#define INF 0x3f3f3f3f
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))

int a1[N], b1[N], c1[N], d1[N], val1[N], val2[N];

int main ()
{
    int a, b, c, d, m, q;

    while (scanf ("%d %d %d %d", &a, &b, &c, &d) != EOF)
    {
        met (a1, 0), met (b1, 0), met (c1, 0), met (d1, 0);
        met (val1, 0), met (val2, 0);

        for (int i=0; i<a; i++) scanf ("%d", &a1[i]);
        for (int i=0; i<b; i++) scanf ("%d", &b1[i]);
        for (int i=0; i<c; i++) scanf ("%d", &c1[i]);
        for (int i=0; i<d; i++) scanf ("%d", &d1[i]);

        int k1 = 0, k2 = 0;

        for (int i=0; i<a; i++)
            for (int j=0; j<b; j++)
                val1[k1++] = a1[i] + b1[j];

        for (int i=0; i<c; i++)
            for (int j=0; j<d; j++)
                val2[k2++] = c1[i] + d1[j];

        scanf ("%d", &m);

        sort (val1, val1+k1);
        sort (val2, val2+k2);

        while (m--)
        {
            int ans = 0;
            scanf ("%d", &q);

            for (int i=0; i<k1; i++)
            {
                int v = q - val1[i];

                int l = lower_bound (val2, val2+k2, v)-val2;///查找val2中>=v的元素下标
                int r = upper_bound (val2, val2+k2, v)-val2;///查找val2中>v的元素下标
                ans += r - l;
            }
            printf ("%d\n", ans);
        }
    }
    return 0;
}