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 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=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…

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:…

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…

#include <string.h> #include <stdio.h> const int maxn = 1000006; bool vis[1000006]; int pr[1000005]; int cnt = 1; int bs(int l, int r, int v) { int mid=(l+r)>>1; while(l < r) { if(pr[mid] < v) l = mid+1; else r = mid; mid= (l+r) &g…

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…

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*/ #…

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]…

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…

Problem 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 conjecture: every even integer can be expressed as the difference of two pr…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4715 思路:先打个素数表,然后判断一下就可以了. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; ]; ]; int main() { ; memset(isprime,true…

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=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 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…

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…

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…

#include<stdio.h> #include<string.h> int prime[1100000],p[1000000],ans; void pri() {     int i,j,k;     memset(prime,-1,sizeof(prime));  prime[0]=prime[1]=0;     for(i=2;i<=1000;i++)         if(prime[i]==-1) {             for(j=i*2;j<=10…

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): 832    Accepted Submission(s): 267 Problem Description All you know Goldbach conjecture.That is to say, Every even inte…

1003 Rotation Lock Puzzle 找出每一圈中的最大值即可 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector> #define ll __int64 #define pi acos(-1.0) #…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5104 Primes Problem Description Given a number n, please count how many tuple$(p_1, p_2, p_3)$ satisfied that $p_1 \leq p_2  \leq p_3, $ $p_1,p_2,p_3$ are primes and $p_1 + p_2 + p_3 = n$. Input Multiple…

Count primes 题目连接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5901 Description Easy question! Calculate how many primes between [1...n]! Input Each line contain one integer n(1 <= n <= 1e11).Process to end of file. Output For each case, output…

http://acm.hdu.edu.cn/showproblem.php?pid=2161 Problem Description Write a program to read in a list of integers and determine whether or not each number is prime. A number, n, is prime if its only divisors are 1 and n. For this problem, the numbers…

Problem Description Writea program to read in a list of integers and determine whether or not eachnumber is prime. A number, n, is prime if its only divisors are 1 and n. Forthis problem, the numbers 1 and 2 are not considered primes. Input Eachinput…

Problem Description Write a program to read in a list of integers and determine whether or not each number is prime. A number, n, is prime if its only divisors are 1 and n. For this problem, the numbers 1 and 2 are not considered primes. Input Each i…