hqd_acm的专栏

做的十分顺利，1Y#include #include long long dp[21][21];double ans[21][21];double out[21][21];long long sum = 0;int main(){ memset(

每次浙大月赛都打酱油，中午这种题，愣是把数字敲错了。。。。。。。#include #include int num[10];char s[20], s2[10];int t[10];int main(){ int n; whil

可能大家都坑在第一题上了，导致这个比较水的题都不做了#include #include #include #include using namespace std;struct Point{ double x, y;} node[201];st

比赛时候脑子堵得慌，注意进位，这两场比赛都发挥的很不好#include #include #include using namespace std;//////////////////////////////////////#define MAX 10000

去年成都赛区的一道题目，画几个图能YY出事凸函数，接着就是三分解#include #include #include using namespace std;const int N = 10002;const double eps = 1e-10;//开成-8就错了

比赛的时候卡题了，这题当时没有看。。。。。题意比较简单，不介绍了简单整理整理思想吧，有些细节的东西是看了标程才明朗起来。题目要求答案是最多收集几个桃子。这种比较容易想到是二分10^4的数据，必须是0(nlgn)以下的算法，二分的话就已经是0(lgn)了，所以下面的验证需要0(n)

北邮邀请赛的题目，比赛的时候思路很乱，现在整理下吧。网上已经给出了有公式解法，我这里还是说说树状数组的解法吧题目只有两个操作，一个是将一个点的颜色取反，一个是询问有多少个端点颜色不同并且包含小于k的区间。总体思路是这样首先是用树状数组保存每个点的颜色，如果是1，那个点加1，这是初

昨天悲剧的卡在这题上了和秦牛推出很多个版本的解法就是蛋疼的不过其实思路是比较简单的就是昨天想的那个关键点，就是把这些方向可以看成力矩一样的东西，也就是矢量吧。如果中间能回到原点，那么后面不管有任何操作，都能回到原点的用一句比较简洁的话概括，就是如果所有的向量都不位于任何一条直线的

这题目的解法果然是很难想到有一个突破点就是任何一个状态，通过题目所给的移动，都能对应且唯一对应一个b*2=a+c(a所以以这个为根如果不是根的状态，可以让左右两个往里跳，依据是让和中间那个坐标距离缩小这样就可以看作为是向根移动具体的操作用的是辗转相除法的思想接着找初始状态和目标状

Sequence AdjustmentTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 477

我是被逼的，做这种鸟递推//HDU 2524 //行列相乘，先处理行的情况，接着把一行看成一个元素#includeint main(){ int n,m,t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); int sum=(n*(n+1))>>1; int k=(m*(m+1))>>1; printf("%d/n",sum*k

<br /> Eeny Meeny MooTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2150 Accepted: 1459<br />DescriptionSurely you have made the experience that when too many people use the Internet simultaneously, the net becomes very, very slow. <br />To put

<br />放苹果Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16904 Accepted: 10656<br />Description把M个同样的苹果放在N个同样的盘子里，允许有的盘子空着不放，问共有多少种不同的分法？（用K表示）5，1，1和1，5，1 是同一种分法。<br />Input第一行是测试数据的数目t（0 <= t <= 20）。以下每行均包含二个整数M和N，以空格分开。1<=M，N<=10。<br />Output对输

<br /> SumsetsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6261 Accepted: 1745<br />DescriptionGiven S, a set of integers, find the largest d such that a + b + c = d where a, b, c, and d are distinct elements of S. <br />InputSeveral S, ea

<br />SumsetsTime Limit: 2000MS Memory Limit: 200000KTotal Submissions: 8611 Accepted: 3461<br />DescriptionFarmer John commanded his cows to search for different sets of numbers that sum to a given number. The cows use only numbers that are an integer

<br />TilingTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5487 Accepted: 2676<br />DescriptionIn how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles? <br />Here is a sample tiling of a 2x17 rectangle. <br /><br />InputInput is a se

<br /> <br /> <br /> <br />//递推#include<cstdio>#include<iostream>#include<string>using namespace std;int dp[30][30];//长度为iint sum[30];void init(){ for(int i=1;i<=26;i++) dp[1][i]=1; sum[1]=26; for(int i=2;i<=10;i++) fo

ParliamentTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12309 Accepted: 5124<br />DescriptionNew convocation of The Fool Land's Parliament consists of N delegates. According to the present regulation delegates should be divided into disjoint gr

Number SequenceTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 22588Accepted: 6019<br />DescriptionA single positive integer i is given. Write a program to find the digit located in the position i in the sequence of number groups S1S2.

Counterfeit DollarTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 27942Accepted: 8742<br />DescriptionSally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit e

PacketsTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 28741Accepted: 9462<br />DescriptionA factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5, 6*6. These products are al

Let it BeadTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2209 Accepted: 1363<br />Description"Let it Bead" company is located upstairs at 700 Cannery Row in Monterey, CA. As you can deduce from the company name, their business is beads. Their P

<br /> Number Sequence<br />Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)<br />Total Submission(s): 37082    Accepted Submission(s): 7790<br />Problem DescriptionA number sequence is defined as follows:<br /><br

<br />A + B Problem IITime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)<br />Total Submission(s): 71329    Accepted Submission(s): 13022<br /><br /><br />Problem DescriptionI have a very simple problem for you. Given

<br /> Temperature<br />Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)<br />Total Submission(s): 393    Accepted Submission(s): 143<br /><br />Problem DescriptionMany people like summer as summer has a lot of advantages

<br />Shell PyramidTime Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)<br />Total Submission(s): 1008    Accepted Submission(s): 364<br /><br /><br />Problem DescriptionIn the 17th century, with thunderous noise, dense smoke

<br />Crazy SearchTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12290 Accepted: 3339<br />DescriptionMany people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a give

Sum of Consecutive Prime NumbersTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9435 Accepted: 5407<br />DescriptionSome positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a

<br /><br />Faulty OdometerTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7042 Accepted: 4375<br />DescriptionYou are given a car odometer which displays the miles traveled as an integer. The odometer has a defect, however: it proceeds from th

<br />大数平方<br />#include<stdio.h>#include<string.h>#include<math.h>int big(char s1[],char s2[]){    int  len1,len2,i,q;    q=0;    while(s1[q]=='0')  q++;    strcpy(s1,s1+q);    if(strlen(s1)==0)    {        s1[0]='0';        s1[1]=0; 

Max Angle Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 12   Accepted Submission(s) : 5Font: Times New Roman | Verdana | GeorgiaFont Size: ←→Problem DescriptionGiven many points in a plane, two player

3904 - Tile CodeAsia - Seoul - 2007/2008PDF Submit RankingThe city of Songpa is now carrying out a project to build a bicycle transportation system called 鈥榞reen Songpa.鈥?By the end of this year, citizens and visitors alike will be able to pick up and drop

<br />很无语<br />#include<stdio.h><br />#include<math.h><br />int main()<br />{<br />    double x,y;<br />    while(scanf("%lf%lf",&x,&y)!=EOF)<br />       printf("%.0lf/n",pow(y,1/x));<br />    return 0;<br />}

<br /><br />UmBasketellaTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3677 Accepted: 1381<br />Description<br />In recent days, people always design new things with multifunction. For instance, you can not only use cell phone to call your fri

<br />FibonacciTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3502 Accepted: 2450<br />Description<br />In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequ

<br /><br />Matrix Power SeriesTime Limit: 3000MS Memory Limit: 131072KTotal Submissions: 5442 Accepted: 2322<br />Description<br />Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.<br />Input<br />The input contains e

<br /><br /> Prime DistanceResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K4617StandardThe branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thous

<br /><br />Pseudoprime numbersTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3171 Accepted: 1105<br />Description<br />Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the 

<br /><br />#include<stdio.h><br />#include<string.h><br />char str[1001];<br />int a[100],b[100];<br />bool isprime[33000];<br />void Make()<br />{<br />       int i,j;<br />       memset(isprime,true,sizeof(isprime));<br />       isprime[0] = isprime[1]

#includeint main(){ int temp,m,n,s;  while(scanf("%d%d",&m,&n)!=EOF)  {  if(m==0&&n==0) break;  if(m==1||n==1) printf("0/n");     else  if(m==2||n==2)  {   if(n!=2) printf("%d/n",n-1);   if(m!=2) prin