HDU 2141 Can you find it?（暴力+二分）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 20473    Accepted Submission(s): 5192

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 31 2 31 2 31 2 331410

 

Sample Output

Case 1:NOYESNO

 

真是醉了，用set本来觉得方法挺好，挺巧，就是爆内存，就因为多开了个数组，malloc都不行。

不用set也可以，开3个数组也不会爆！

大体题意：

输入3组数组，问是否能在3组数组中找出三个数来，使得他们的和是X

思路：

三层循环肯定爆，所以可以枚举前两个数组和的所有情况，输入一个数，枚举C数组，用二分查找那个数组和的数组。

代码如下：

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cstdlib>#include<set>using namespace std;const int maxn = 500 + 10;int L,M,N,cnt=0,cnt2;int a[maxn],b[maxn*maxn],c[maxn],ok;bool my_find(int k){    int l=0,r=cnt2-1;    while(l <= r){        int mid = l + (r - l) / 2;        if (b[mid] == k)return true;        if (b[mid] < k)l = mid+1;        else r = mid-1;    }    return false;}int main(){    while(cin >> L >> M >> N){        cnt2 = 0;        for (int i = 0; i < L; ++i)scanf("%d",&a[i]);        for (int i = 0; i < M; ++i){            int h;            scanf("%d",&h);            for (int j = 0; j < L; ++j)b[cnt2++]=a[j]+h;        }        for (int i = 0; i < N; ++i)scanf("%d",&c[i]);        sort(c,c+N);        sort(b,b+cnt2);        int cont,sum;        cin >> cont;        printf("Case %d:\n",++cnt);        while(cont--){            ok=false;            cin >> sum;            for (int i = 0; i < N; ++i){                int k = sum - c[i];                //if (k < 0)goto TT;                if (my_find(k)){                    ok=true;                    break;                }            }                if (ok)printf("YES\n");                else printf("NO\n");        }    }    return 0;}