HDU 4981 Goffi and Median 思路:排序就能够得到中间数.然后总和和中间数*n比較一下就可以 代码: #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int N = 1005; int n, a[N], sum; int main() { while (~scanf("%d…

题解:排序取中位数,然后与平均数比较即可. #include <cstdio> #include <algorithm> using namespace std; double a[1005],ave,med,sum; int n; int main(){ while(~scanf("%d",&n)){ sum=0; for(int i=1;i<=n;i++){scanf("%lf",&a[i]);sum+=a[i];}…

HDU 4982 Goffi and Squary Partition 思路:直接从全然平方数往下找,然后推断是否能构造出该全然平方数,假设能够就是yes,假设都不行就是no.注意构造时候的推断,因为枚举一个全然平方数.剩下数字为kk.构造的时候要保证数字不反复 代码: #include <cstdio> #include <cstring> #include <cmath> int n, k; bool judge(int num) { int yu = num *…

题目链接:hdu 4983 Goffi and GCD 题目大意:求有多少对元组满足题目中的公式. 解题思路: n = 1或者k=2时:答案为1 k > 2时:答案为0(n≠1) k = 1时:须要计算,枚举n的因子.令因子k=gcd(n−a,n, 那么还有一边的gcd(n−b,n)=nk才干满足相乘等n.满足k=gcd(n−a,n)的a的个数即为ϕ(n/s),欧拉有o(n‾‾√的算法 #include <cstdio> #include <cstring> #include…

HDU 4983 Goffi and GCD 思路:数论题.假设k为2和n为1.那么仅仅可能1种.其它的k > 2就是0种,那么事实上仅仅要考虑k = 1的情况了.k = 1的时候,枚举n的因子,然后等于求该因子满足的个数,那么gcd(x, n) = 该因子的个数为phi(n / 该因子),然后再利用乘法原理计算就可以 代码: #include <cstdio> #include <cstring> #include <cmath> typedef long lo…

Friendship of Frog Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5578 Description N frogs from different countries are standing in a line. Each country is represented by a lowercase letter. The distance betwee…

ZYB's Biology Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5590 Description ZYB(ZJ−267)在NOIP拿到600分之后开始虐生物题,他现在扔给你一道简单的生物题:给出一个DNA序列和一个RNA序列,问它们是否配对. DNA序列是仅由A,C,G,T组成的字符串,RNA序列是仅由A,C,G,U组成的字符串. DNA和RNA匹配当且仅当每…

L - House Building Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5538 Description Have you ever played the video game Minecraft? This game has been one of the world's most popular game in recent years. The wor…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4813 签到题. 把一个字符串按照格式输出就可以了,很水 #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <queue> #include <map> #include <set> #include…

HDU 4847 Wow! Such Doge! pid=4847" style="">题目链接 题意:给定文本,求有几个doge,不区分大写和小写 思路:水题.直接一个个读字符每次推断就可以 代码: #include <stdio.h> #include <string.h> char c; char a[5]; int main() { a[5] = '\0'; int ans = 0; while ((c = getchar()) != E…

http://acm.hdu.edu.cn/showproblem.php?pid=5104 找元组数量,满足p1<=p2<=p3且p1+p2+p3=n且都是素数 不用素数打表都能过,数据弱的一比 #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <queue> #inclu…

Goffi and Squary Partition Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Recently, Goffi is interested in squary partition of integers. A set X of k distinct positive i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1062 解题报告:注意一行的末尾可能是空格,还有记得getchar()吃回车符. #include<cstdio> #include<string.h> #include<iostream> #include<algorithm> #include<cmath> #include<deque> #include<cstdlib>…

Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 89984    Accepted Submission(s): 21437 Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A…

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21106    Accepted Submission(s): 9498 Problem Description In many applications very large integers numbers are required. Some of these…

超级楼梯 Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description 有一楼梯共M级,刚开始时你在第一级,若每次只能跨上一级或二级,要走上第M级,共有多少种走法? Input 输入数据首先包含一个整数N,表示测试实例的个数,然后是N行数据,每行包含一个整数M(<=M<=),表示楼梯的级数. Outpu…

其实如果想出了方法真的好水的说... 然而一开始想了好久都没想出来... 最后看了一下最大数据才32768 可以直接枚举...枚举每个硬币的数量 看看后来能不能凑够n 因为还是怕超时..(虽然只有3乘十的四次方)所以先枚举三分 然后二分 最后一分肯定足够 就不用了 自己好傻... #include<stdio.h> #include<string.h> #include<algorithm> #include<map> #include<math.h&…

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…

ZYB loves Score Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5268 Description One day,ZYB participated in the BestCoder Contest There are four problems. Their scores are 1000,1500,2000,2500 According to the r…

http://acm.hdu.edu.cn/showproblem.php?pid=4707 [题目大意]: Lin Ji 的宠物鼠丢了,在校园里寻找,已知Lin Ji 在0的位置,输入N D,N表示校园中点的个数,D表示宠物鼠不可能在距离D之内,接下来N-1行,输入x,y,表示x与y相邻,(相邻两点之间的距离为1,不相邻为inf),不存在环结构. [题解]: 用邻接表存储树形结构,然后用DFS遍历图 [code]: #include <iostream> #include <stdio…

A Sweet Journey Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5477 Description Master Di plans to take his girlfriend for a travel by bike. Their journey, which can be seen as a line segment of length L, is a r…

A Curious Matt Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5112 Description There is a curious man called Matt. One day, Matt's best friend Ted is wandering on the non-negative half of the number line. Matt f…

Problem Description You may have heard of the book '2001 - A Space Odyssey' by Arthur C. Clarke, or the film of the same name by Stanley Kubrick. In it a spaceship is sent from Earth to Saturn. The crew is put into stasis for the long flight, only tw…

Problem Description 妈妈每天都要出去买菜,但是回来后,兜里的钱也懒得数一数,到底花了多少钱真是一笔糊涂帐.现在好了,作为好儿子(女儿)的你可以给她用程序算一下了,呵呵. Input 输入含有一些数据组,每组数据包括菜种(字串),数量(计量单位不论,一律为double型数)和单价(double型数,表示人民币元数),因此,每组数据的菜价就是数量乘上单价啊.菜种.数量和单价之间都有空格隔开的. Output 支付菜价的时候,由于最小支付单位是角,所以总是在支付的时候采用四舍五入的…

Problem Description 8600的手机每天消费1元,每消费K元就可以获赠1元,一开始8600有M元,问最多可以用多少天? Input 输入包括多个测试实例.每个测试实例包括2个整数M, k,(2 <= k <= M <= 1000).M = 0, k = 0代表输入结束. Output 对于每个测试实例输出一个整数,表示M元可以用的天数. Sample Input 2 2 4 3 0 0 Sample Output 3 5 水题.... import java.util.…

Problem Description Goffi is doing his math homework and he finds an equality on his text book: gcd(n−a,n)×gcd(n−b,n)=nk. Goffi wants to know the number of (a,b) satisfy the equality, if n and k are given and 1≤a,b≤n. Note: gcd(a,b) means greatest co…

HDU 4931 Happy Three Friends 题目链接 题意:6个数字,一个取两个,妹子取三个,问最后谁会赢 思路:排个序,推断前两个和3 - 5个的和谁更大就可以 代码: #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; int t, a[6]; int main() { scanf("%d&q…

已知圆心(0,0)圆周上的一点,求圆周上另外两点使得三点构成等边三角形. 懒得推公式,直接用模板2圆(r1=dist,r2=sqrt(3)*dist)相交水过 #include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<iterator> using namespace std; #define eps 1e-6 typedef long lon…

本来是不打算贴这道水题的,自己却WA了三次.. 要考虑1的情况,1的质因子为1 思路:先打表 ,然后根据最大质因子更新结果 代码: #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> using namespace std; #define MAX 20000 int p[MAX]; int main() { memset(p,,sizeof(p)); p[]=; ;…

Equations 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1840 ——每天在线,欢迎留言谈论. 题目大意: 给你一个一元二次方程组,a(X^2) + b(X) + c = 0 .求X解的个数. 思路: 分别讨论二次方程与一次方程的情况,再特殊处理下 a = b = c = 0 的情况. 感想: 是时候该水水题了. Java AC代码: import java.math.*; import java.util.Scanner; public…