Trouble Time Limit: 5000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status Description Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him. The 5-sum problem is def…

Trouble Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3526    Accepted Submission(s): 1113 Problem Description Hassan is in trouble. His mathematics teacher has given him a very difficult pro…

Trouble Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5681    Accepted Submission(s): 1570 Problem Description Hassan is in trouble. His mathematics teacher has given him a very difficult pro…

Trouble Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5388    Accepted Submission(s): 1494 Problem Description Hassan is in trouble. His mathematics teacher has given him a very difficult pro…

给定五个集合.问是否能从五个集合各取一个元素,使得元素之和为0. 这道题有两种做法,一种是哈希,然而之前没写过哈希.....比赛后从大神那copy了一份. 这里说还有一种. 对于这五个集合分为三组.1,2组求和为一组,3,4组求和分为一组,5为一组. 那么如今转化为了是否能从前两组中各取一个元素.使得这两个值和为第三组一个元素的相反数. 那么对于第一组我们升序排序.第二组我们降序排序. 对于第三组里的任一元素,假如第一组队首加第二组队首之和大于第三组的元素.那么第二组游标往后移一位,反之第一组移…

[hdu P4334] Trouble Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him. The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,.…

Hassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him. The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such t…

题目:click here 题意: 给定5组数据,每组数据选择一个数,看是否能找到5个数的和为零. 分析: 千万不要~~T~~ 普通线性查找: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long ll; const int INF = 0x3f3f3f3f; ; int t, n…

原题链接 思路:哈希存入相反数 注意:HDU不支持long long要使用__int64 #include<cstdio> #include<cstring> #define mod 401704//开到所有数据十倍大左右 using namespace std; bool flag[mod]; __int64 m[][],hash[mod]; int judge(__int64 x) { int s=x%mod; ) s+=mod;//将负数转正 /*while(x<0)…

题目大意: 从5个集合中个选取一个数出来,使5个数相加之和为0 , 判断是否存在这种可能 因为集合数目最多200,那么200^3 = 8000000 , 那么这里很明显要把5个数拆成2个和3个计算,因为3个的话有8000000种可能,不好保存 所以只先算前两个数40000种相加的可能性保存到hash表中,然后再后面800000次找hash表中是否存在对应的值 #include <cstdio> #include <cstring> #include <iostream>…