Difference Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 581    Accepted Submission(s): 152 Problem Description A graph is a difference if every vertex vi can be assigned a real number ai and…

这是通化邀请赛的题,当时比赛的时候还完全没想法呢,看来这几个月的训练还是有效果的... 题意要求(1) |ai| < T for all i   (2) (vi, vj) in E <=> |ai - aj| >= T.由于(1)条件的存在,所以(2)条件能成立当且仅当ai和aj一正一负.由此可见,图中某条路上的元素正负值分别为正->负->正->负...显然当图中存在奇环的时候是无解的.判断奇环用二分染色,color[i]=0表示假设i节点未被染色,1表示假设i节…

HDU 5487 Difference of Languages 这题从昨天下午2点开始做,到现在才AC了.感觉就是好多题都能想出来,就是写完后debug很长时间,才能AC,是不熟练的原因吗?但愿孰能生巧吧. BFS转移的是两个DFA的状态,用typedef pair<int,int> pi;map<pi,pi> pres;      两步储存前后状态的链接. 附上一组测坑数据: /*4324 3 130 1 a1 2 e2 3 w6 3 150 1 a1 3 e3 5 h*/ #…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4715 Difference Between Primes Description All you know Goldbach conjecture.That is to say, Every even integer greater than 2 can be expressed as the sum of two primes. Today, skywind present a new conje…

Difference Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 62    Accepted Submission(s): 19 Problem Description Little Ruins is playing a number game, first he chooses two positive integers y an…

Difference of Languages Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 548764-bit integer IO format: %I64d      Java class name: Main A regular language can be represented as a deterministic finite automaton…

Difference of Clustering HDU - 5486 题意:有n个实体,新旧两种聚类算法,每种算法有很多聚类,在同一算法里,一个实体只属于一个聚类,然后有以下三种模式. 第一种分散,新算法的某几个聚类是旧算法某个聚类的真子集. 第二种聚合,旧算法的某几个聚类是新算法某个聚类的真子集. 第三种1:1,新算法的某个聚类跟旧算法某个聚类相同. 问每种模式存在多少个? 实路上很明了,就是模拟,而且每个实体最多遍历到一次,所以时间上不用担心. 先把聚类的编号hash,把每个实体分配到相应…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5486 题意: 给你每个元素一开始所属的集合和最后所属的集合,问有多少次集合的分离操作,并操作和不变操作. 分离:[m1,m2,m3]->[m1],[m2],[m3] 合并:分离的逆操作 不变:[m1,m2,m3]->[m1,m2,m3] 题解: 以集合为单位建图,(一个元素从集合s1到s2则建一条边连接集合s1,s2,注意要删除重边) 然后对于每个点,与它相邻的点如果入度都为1,则为分离操作,…

http://acm.hdu.edu.cn/showproblem.php?pid=4715 [code]: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; #define N 1000151 ]; ]; ]; ]; ; int lowbit(int i) { return i&-i; } void add(i…

Difference Between Primes Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 860 Accepted Submission(s): 278 Problem Description All you know Goldbach conjecture.That is to say, Every even integer gr…

Difference Between Primes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 528    Accepted Submission(s): 150 Problem Description All you know Goldbach conjecture.That is to say, Every even inte…

http://acm.hdu.edu.cn/showproblem.php?pid=5936 题意: 定义了这样一种算法,现在给出x和k的值,问有多少个y是符合条件的. 思路: y最多只有10位,再多x就是负的了. 这样的话可以将y分为前后两部分,我们先枚举后5位的情况,然后再枚举前5位的情况,通过二分查找找到匹配的项,这样就大大的降低了时间复杂度. #include<iostream> #include<algorithm> #include<cstring> #in…

Difference of Clustering Problem Description Given two clustering algorithms, the old and the new, you want to find the difference between their results. A clustering algorithm takes many member entities as input and partition them into clusters. In…

题意: 有一个函数f(y, k) = y的每个十进制位上的数字的k次幂之和 给x, k 求 有多少个y满足 x = f(y, k) - y 思路: (据说这叫中途相遇法?) 由于 x >= 0 所以 显然y最多也不会超过10位数 把一个数拆成前5位 和 后5位 即找有多少对(a, b)满足 x = a + b 把所有的后五位预处理出来,然后再找前五位. 找的时候用二分,mapT了..可能组数太多了吧. 具体看代码 + ; LL num[maxn]; int x, k; LL p[][]; LL…

思路:首先就是判断是否有奇环,若存在奇环,则输出No. 然后用差分约束找是否符合条件. 对于e(i,j)属于E,并且假设顶点v[i]为正数,那么v[i]-v[j]>=T--->v[j]-v[i]<=-T; 对于e(i,j)不属于E,并且假设顶点v[i]为正数,那么v[i]-v[j]<=T-1; 以0号点为参照点,若果v[i]为负数,那么v[0]-v[i]<=T-1;  v[i]-v[0]<=0; 若果v[i]为正数,那么v[i]-v[0]<=T-1;  v[0]-…

题意:给出一个偶数(不论正负),求出两个素数a,b,能够满足 a-b=x,素数在1e6以内. 只要用筛选法打出素数表,枚举查询下就行了. 我用set储存素数,然后遍历set里面的元素,查询+x后是否还是素数. 注意,偶数有可能是负数,其实负数就是将它正数时的结果颠倒就行了. 代码: /* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.csdn.net/hcbbt * File: 10.cpp * Create Date:…

这道题其实不需要考虑具体数值,但可以肯定的是,相连边的两端点必定有一正一负,至于谁正谁负,并不重要,这是可以思考的,很明显的一个二分图性质,如果不满足此条件,是不可能满足题目第二个条件的.所以首先对题目二分染色.对于T,随意指定一个值即可. 注意题目第二个条件为充要条件,所以要考虑没有相连的两点. uv相连时,设u是正数,很明显有u-v<=-T uv不连时,设u是正数,有v-u<=T-1. 对于第一个条件,虚拟一个源点即可.注意对于某点u(正或负),判断它和零的关系. 我的代码,未A,查不出错…

题目链接 题意 : 给出一个 x 和 k 问有多少个 y 使得 x = f(y, k) - y .f(y, k) 为 y 中每个位的数的 k 次方之和.x ≥ 0 分析 : f(y, k) - y = x ≥ 0 满足条件的 y 最多不超过 10 位 这个并不知道怎么证.网上有很多结论证明.仔细推敲过后发现都是错的 可能需要手动打表模拟一下吧...... 变化一下式子 x = f(a, k) - a + f(b, k) - b * (1e5) x - f(a, k) + a = f(b, k)…

Difference Between Primes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3339    Accepted Submission(s): 953 Problem Description All you know Goldbach conjecture.That is to say, Every even int…

Iterated Difference Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 786    Accepted Submission(s): 505 Problem Description You are given a list of N non-negative integers a(1), a(2), ... , a(N).…

Little Ruins is playing a number game, first he chooses two positive integers yy and KK and calculates f(y,K)f(y,K), here f(y,K)=∑z in every digits of yzK(f(233,2)=22+32+32=22)f(y,K)=∑z in every digits of yzK(f(233,2)=22+32+32=22) then he gets the re…

这道题很坑,注意在G++下提交,否则会WA,还有就是a或b中较大的那个数的范围.. #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int maxn = 1e6 + 10; int prime[maxn]; bool isprime[maxn]; int init() { memset(prime, 0, sizeof(prime)); isprime[0]…

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Nim or not Nim? Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 3032 Description Nim is a two-player mathematic game of strategy in which players take turns removing objects from distinct heaps.…

zxa and leaf Problem Description zxa have an unrooted tree with n nodes, including (n−1) undirected edges, whose nodes are numbered from 1 to n. The degree of each node is defined as the number of the edges connected to it, and each node whose degree…

http://acm.hdu.edu.cn/showproblem.php?pid=4715 Difference Between Primes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description All you know Goldbach conjecture.That is to say, Every even integer great…

http://acm.split.hdu.edu.cn/showproblem.php?pid=4336 Card Collector Special Judge Problem Description   In your childhood, do you crazy for collecting the beautiful cards in the snacks? They said that, for example, if you collect all the 108 people i…

http://acm.hdu.edu.cn/showproblem.php?pid=1700 Points on Cycle Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1567    Accepted Submission(s): 570 Problem Description There is a cycle with its c…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3530 Subsequence Description There is a sequence of integers. Your task is to find the longest subsequence that satisfies the following condition: the difference between the maximum element and the minim…