HDU Problem 2141 Can you find it? 【二分】

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 23856 Accepted Submission(s): 6048

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

Sample Output

Case 1:
NO
YES
NO

#include <cstdio>
#include <cmath>
#include <algorithm>
#define MAX_N 500050
using namespace std;
const double ESP = 1e-5;
const int INF = 1e8;
int L[MAX_N], N[MAX_N], M[MAX_N], LN[MAX_N];
int M1[MAX_N];
int l, n, m, cnt;

bool lower_bound(int x) {
    int lp = -1, rp = cnt - 1;
    while (rp - lp > 1) {
        int mid = (lp + rp)/2;
        if (LN[mid] >= x) rp = mid;
        else lp = mid;
    }
    //printf("%d %d
", x, LN[rp]);
    return LN[rp] == x;
}

int main() {
    int t, p;
    int cut = 0;
    while (scanf("%d%d%d", &l, &n, &m) != EOF) {
        printf("Case %d:
", ++cut);
        for (int i = 0; i < l; i++)
            scanf("%d", &L[i]);
        for (int i = 0; i < n; i++)
            scanf("%d", &N[i]);
        for (int i = 0; i < m; i++)
            scanf("%d", &M1[i]);
        //利用等式L[i]+N[i] = x - M[i]降低运行时间
        cnt = 0;
        for (int i = 0; i < l; i++) {
            for (int j = 0; j < n; j++) {
                LN[cnt++] = L[i] + N[j];
            }
        }
        sort(LN, LN + cnt);
        //printf("%d
", LN[0]);
        scanf("%d", &t);
        while (t--) {
            bool flag = false;
            scanf("%d", &p);
            //用p减去M中的每一个元素
            for (int i = 0; i < m; i++) {
                M[i] = p - M1[i];
            }
            sort(M, M + m);
            for (int i = 0; i < m; i++) {
                if (lower_bound(M[i])) {
                    flag = true;
                    break;
                }
            }
            if (flag)   printf("YES
");
            else printf("NO
");
        }
    }
    return 0;
}