HDU - 2141 Can you find it?

题意：从序列A，B，C中分别选一个数，有无可能等于X。

分析：

1、枚举A中所有数，由此可知X-Ai。

2、统计记录Bi+Ci的所有可能性

3、在 这所有可能性里二分找X-Ai

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 500 + 10;
const int MAXT = 250000 + 10;
using namespace std;
int sum[MAXT];
int a[MAXN];
int b[MAXN];
int c[MAXN];
int cnt;
bool solve(int x){
    int l = 0, r = cnt - 1;
    while(l <= r){
        int mid = l + (r - l) / 2;
        if(sum[mid] == x) return true;
        else if(sum[mid] < x) l = mid + 1;
        else r = mid - 1;
    }
    return false;
}
int main(){
    int L, M, N;
    int kase = 0;
    while(scanf("%d%d%d", &L, &M, &N) == 3){
        memset(sum, 0, sizeof sum);
        memset(a, 0, sizeof a);
        memset(b, 0, sizeof b);
        memset(c, 0, sizeof c);
        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]);
        }
        cnt = 0;
        for(int i = 0; i < M; ++i){
            for(int j = 0; j < N; ++j){
                sum[cnt++] = b[i] + c[j];
            }
        }
        sort(sum, sum + cnt);
        int S;
        scanf("%d", &S);
        printf("Case %d:\n", ++kase);
        while(S--){
            bool ok = false;
            int x;
            scanf("%d", &x);
            for(int i = 0; i < L; ++i){
                int t = x - a[i];
                if(solve(t)){
                    printf("YES\n");
                    ok = true;
                    break;
                }
            }
            if(!ok) printf("NO\n");
        }
    }
    return 0;
}