/* HDU 6168 - Numbers [ 思维 ] | 2017 ZJUT Multi-University Training 9 题意: .... 分析: 全放入multiset 从小到大,慢慢筛 */ #include <bits/stdc++.h> using namespace std; const int N = 125250; int n, s[N]; int a[N], cnt; multiset<int> st; multiset<int>::it…

转载请注明出处:http://blog.csdn.net/a1dark 分析:一道水题.找下规律就OK了.不过要注意特判一下0.因为0也是good number.这个把我坑惨了= =||| #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> using namespace std; int main(){ __int64 a,b; int t,step=1; scan…

Numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5522 Description 给n个数A1,A2....An{A}_{1},{A}_{2}....{A}_{n}A​1​​,A​2​​....A​n​​,从中选3个位置不同的数A,B和C,问是否有一种情况满足A-B=C. Input There are multiple test cases, no mo…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5585 思路:对于2和5只须看最后一位数,对于三看所有位的数字之和就行 #include<stdio.h> #include<math.h> #include<stdlib.h> #include<string.h> #include<iterator> #include<iostream> #include<algorithm>…

Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 460    Accepted Submission(s): 283 Problem Description There is a number N.You should output "YES" if N is a multiple of 2, 3 or 5,ot…

http://acm.hdu.edu.cn/showproblem.php?pid=5181 题意: 有一个栈,其中有n个数1~n按顺序依次进入栈顶,在某个时刻弹出. 其中m个限制,形如数字A必须在数字B之前弹出. 求方案总数   dp[i][j]表示数字i~j的出栈方案数 枚举最后一个出栈的数k,若k合法 dp[i][j]+=dp[i][k]*dp[k+1][j] 如何判断k是否合法?   对于一组i,j,k来说,它的弹出顺序是 [i,k-1]早于[k+1,j]早于k 对于一个限制 A必须在B…

Numbers Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 514    Accepted Submission(s): 270 Problem Description zk has n numbers a1,a2,...,an. For each (i,j) satisfying 1≤i<j≤n, zk generates a…

题目链接 Problem Description zk has n numbers a1,a2,...,an. For each (i,j) satisfying 1≤i<j≤n, zk generates a new number (ai+aj). These new numbers could make up a new sequence b1,b2,...,bn(n−1)/2. LsF wants to make some trouble. While zk is sleeping, Ls…

Problem Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers. Write a program to find and pri…

Humble Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21853    Accepted Submission(s): 9553 Problem Description A number whose only prime factors are 2,3,5 or 7 is called a humble numb…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=5181 题解:https://www.cnblogs.com/Miracevin/p/10960717.html 原来卡特兰数的这个问题还能区间DP…… XO #include<cstdio> #include<cstring> #include<algorithm> #define ll long long using namespace std; int rdn() { ;;…

水题 #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; ]; int n; ]; int main() { while(~scanf("%d",&n)){ memset(tot,,sizeof tot); ;i<=n;i++) { scanf("%d",&a[i])…

题目链接 参考. \(Description\) 将\(1,2,\cdots,n(n\leq 300)\)依次入栈/出栈,并满足\(m(m\leq 90000)\)个形如\(x\)要在\(y\)之前出栈的限制,问合法的出栈序列有多少种. \(Solution\) 没有限制就是个卡特兰数,但有了限制就要考虑好好DP了.. 序列的入栈&出栈顺序可以构成一棵二叉树,且每一棵子树中的点一定比该子树的根节点出栈早. \(f[i][j]\)表示子树根节点为\(i\),其中的点是\(i\sim j\),\(i…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5272 Dylans loves numbers Description Who is Dylans?You can find his ID in UOJ and Codeforces.His another ID is s1451900 in BestCoder. And now today's problems are all about him. Dylans is given a number…

Dylans loves numbers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5272 Description Dylans是谁?你可以在 UOJ 和 Codeforces上看到他.在BestCoder里,他有另外一个ID:s1451900.今天的题目都和他有关哦.Dylans得到了一个数N.他想知道N的二进制中有几组1.如果两个1之间有若干个(至少一个)0…

http://acm.hdu.edu.cn/showproblem.php?pid=2817 __int64 pow_mod (__int64 a, __int64 n, __int64 m)快速幂取模函数. A sequence of numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4047    Accepted Su…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4722 Good Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 422    Accepted Submission(s): 146 Problem Description If we sum up every digit o…

原题直通车: HDU  4722  Good Numbers 题意: 求区间[a,b]中各位数和mod 10==0的个数. 代码: #include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; int f[20]; long long work(long long x){ long long ret=0, u=x; int t=0, s=0…

题目链接: PKU:http://poj.org/problem?id=3340 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2410 Description A wild number is a string containing digits and question marks (like 36?1?8). A number X matches a wild number W if they have the same length, and…

HDU 3117 Fibonacci Numbers(斐波那契前后四位,打表+取对+矩阵高速幂) ACM 题目地址:HDU 3117 Fibonacci Numbers 题意:  求第n个斐波那契数的前四位和后四位.  不足8位直接输出. 分析:  前四位有另外一题HDU 1568,用取对的方法来做的.  后四位能够用矩阵高速幂,MOD设成10000即可了. 代码: /* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.c…

Humble Numbers Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1058 Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9…

HDOJ(HDU).1058 Humble Numbers (DP) 点我挑战题目 题意分析 水 代码总览 /* Title:HDOJ.1058 Author:pengwill Date:2017-2-15 */ #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #define nmax 6000 #define min(a,b) (a<b?a:b…

A - Arcane Numbers 1 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4320 Description Vance and Shackler like playing games. One day, they are playing a game called "arcane numbers". Th…

2019 杭电多校 8 1003 题目链接:HDU 6659 比赛链接:2019 Multi-University Training Contest 8 Problem Description Acesrc is a famous mathematician at Nanjing University second to none. Playing with interesting numbers is his favorite. Today, he finds a manuscript whe…

斐波那契数列后四位可以用快速幂取模(模10000)算出.前四位要用公式推 HDU 3117 Fibonacci Numbers(矩阵快速幂+公式) f(n)=(((1+√5)/2)^n+((1-√5)/2)^n)/√5 假设F[n]可以表示成 t * 10^k(t是一个小数),那么对于F[n]取对数log10,答案就为log10 t + K,此时很明显log10 t<1,于是我们去除整数部分,就得到了log10 t 再用pow(10,log10 t)我们就还原回了t.将t×1000就得到了F[n…

题目链接:hdu 5676 一开始看题还以为和数位dp相关的,后来才发现是搜索题,我手算了下,所有的super lucky number(也就是只含数字4, 7且4, 7的数量相等的数)加起来也不过几万个,可以采用打表的方法来把所有的super lucky number存储起来.因为4,7数量须相等,所以可以用一个二进制数的0,1来代替,先限定4,7数量分别为 i,之后就是求出包含 i 个0和 i 个1的 2*i 位所有这样的二进制数,然后简单转换一下(1->7, 0->4,这样子能从小到大…

Quite Good Numbers Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB Total submit users: 77, Accepted users: 57 Problem 12876 : No special judgement Problem description A "perfect" number is an integer that is equal to the sum…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1058 解题报告:输入一个n,输出第n个质因子只有2,3,5,7的数. 用了离线打表,因为n最大只有5842. #include<stdio.h> #define INT __int64 INT ans[] = { ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5108 题目意思:给出一个数正整数 N,N <= 1e9,现在需要找出一个最少的正整数 M,使得 N/M 是素数.如果找不到就输出0. 一开始有想过将所有 <= 1e9 的素数求出来的,不过绝对超时就放弃了:然后就开始从题目中挖掘简便的处理方法.受到求素数的方法启发,枚举的因子 i 如果在 i * i <= N 之内都没有找到符合条件的素数,那么那些 > N 的因子就更不可能了.于是时间…

Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers. Now given a humble number, please write…