hdoj2141 Can you find it?

题目：

Problem Description
Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.
Input
There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.
Output
For each case, firstly you have to print the case number as the form “Case d:”, then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print “YES”, otherwise print “NO”.
Sample Input
3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10
Sample Output
Case 1:
NO
YES
NO

这道题是给你3个数字，分别是3个数组元素的个数，然后是3个数组。之后给你一个数字，代表要测试多少组，每一组一个数字，如果能分别从三个数组中各选出一个，使这三个数字之和等于该数字，则为YES，否则为NO.

如果直接三重循环，肯定超时，因此此题需要对方法进行优化。

此题可以先用二重循环将前面两个数组的和的所有情况列出，单独存入一个数组。之后遍历另一个数组，将测试数字与另一个数组里面的数字相减，看结果能否在那个由前两个数组元素之和的数组内找到。

找数字可以用二分的方法进行优化。

参考代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>const int N = 500+10;using namespace std;int l,m,n;int k;int a[N],b[N],c[N];int sum[N*N];void init() {    memset(a,0,sizeof(a));    memset(b,0,sizeof(b));    memset(c,0,sizeof(c));    memset(sum,0,sizeof(sum));}void input() {    for (int i = 0;i < l;++i) {        scanf("%d", &a[i]);    }    for (int i = 0;i < m;++i) {        scanf("%d", &b[i]);    }    for (int i = 0;i < n;++i) {        scanf("%d", &c[i]);    }}void combin() {    k = 0;    for (int i = 0;i < l;++i) {        for (int j = 0;j < m;++j) {            sum[k++] = a[i] + b[j];         }    }    sort(sum,sum+k);}void twopoint(int num) {    bool flag = false;    for (int i = 0;i < n;++i) {        int key = num - c[i];        int left = 0,right = k - 1;        while (left <= right) {            int mid = left + (right - left) / 2;            if (sum[mid] == key) {                flag = true;                break;            }            else if (key < sum[mid]) {                right = mid - 1;            }            else {                left = mid + 1;            }        }    }    if (flag) printf("YES\n");    else printf("NO\n");    } int main() {    int cnt = 1;    while (scanf("%d%d%d", &l, &m, &n) != EOF) {        init();        input();        int s;        int num;        scanf("%d", &s);        combin();        printf("Case %d:\n", cnt);                for (int i = 1;i <= s;++i) {            scanf("%d", &num);            twopoint(num);            }        cnt++;    }    return 0;}

另外，此题也有需要注意的地方：先求和（类似打表），不然也有可能超时。
此题的输入输出也需要注意。