hdu 2141

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Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 11439    Accepted Submission(s): 3014

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

题意：二分。Ai+Bj+Ck = X.所以Ai+Bj=X-Ck

这样就可以用二分了，用一个数组将Ai+Bj存起来，然后在一步步的去查找X-Ck。

这里面有所有的二分全集，这个在时间优化上比较性点击打开链接。

#include <iostream>#include <algorithm>#include<cstdio>using namespace std;const int N = 505;__int64 ab[N * N];int num;int search(__int64 x){int f = 0, l = num - 1;int mid;while(f <= l){mid = (f + l) / 2;if(ab[mid] == x)return 1;else if(ab[mid] < xf = mid + 1;elsel = mid - 1;}return 0;}int main(){int n, m, l, flag = 0, s;__int64 a[N], b[N], c[N], x;while(cin >> n >> m >> l){flag++;num = 0;for(int i = 0; i < n; i++)scanf("%I64d", &a[i]);for(int i = 0; i < m; i++)scanf("%I64d", &b[i]);for(int i = 0; i < l; i++)scanf("%I64d", &c[i]);for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)ab[num++] = a[i] + b[j];sort(ab, ab+num);sort(c, c+l);scanf("%d", &s);printf("Case %d:\n", flag);while(s--){scanf("%I64d", &x);if(x < ab[0] + c[0] || x > ab[num-1] + c[l-1])printf("NO\n");else{__int64 p;int j;for(j = 0; j < l; j++){p = x - c[j];if(search(p)){printf("YES\n");break;}}if(j == l)printf("NO\n");}}}return 0;}