杭电2034人人都爱A-B#include#include#include#define N 102int main()

杭电acm 2072 求错在哪里了.输出来时全是1.
单词数
Problem Description
lily的好朋友xiaoou333最近很空,他想了一件没有什么意义的事情,就是统计一篇文章里不同单词的总数.下面你的任务是帮助xiaoou333解决这个问题.
Input
有多组数据,每组一行,每组就是一篇小文章.每篇小文章都是由小写字母和空格组成,没有标点符号,遇到#时表示输入结束.
Output
每组只输出一个整数,其单独成行,该整数代表一篇文章里不同单词的总数.
Sample Input
you are my friend
#
Sample Output
4
#include "stdio.h"
#include "string.h"
int main()
{
x05char str[100];
x05int i,word=0,l;
x05while (scanf("%s",str)!=EOF)
x05{
x05x05if (str[0]=='#')
x05x05x05break;
x05x05word=0;
x05x05l=strlen(str);
x05x05if ((str[0]>='a' && str[0]='A' && str[0]

杭电acm 2034
wrong answer
Problem Description
参加过上个月月赛的同学一定还记得其中的一个最简单的题目,就是{A}+{B},那个题目求的是两个集合的并集,今天我们这个A-B求的是两个集合的差,就是做集合的减法运算.（当然,大家都知道集合的定义,就是同一个集合中不会有两个相同的元素,这里还是提醒大家一下）
Input
每组输入数据占1行,每行数据的开始是2个整数n(0

A+B for Input-Output Practice (VIII) 为什么 我在 杭电 上 做是 Presentation Error!
Problem Description
Your task is to calculate the sum of some integers.
Input
Input contains an integer N in the first line,and then N lines follow.Each line starts with a integer M,and then M integers follow in the same line.
Output
For each group of input integers you should output their sum in one line,and you must note that there is a blank line between outputs.
Sample Input
3
4 1 2 3 4
5 1 2 3 4 5
3 1 2 3
Sample Output
10
15
6
Author
lcy
Recommend
JGShining
#include
int main( )
{
x05int a,b,i,j,m,n;
while(scanf("%d",&n)!=EOF)
x05{
x05x05for(j=0;j

杭电ACM FatMouse' Trade算法题
Problem Description
FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input
The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

Output
For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.
Sample Input
5 37 24 35 2-1 -1
Sample Output
13.333

我的代码,请问哪里有错啊,谢谢
#include
int main(){
int m,i,j,k;
float rate,sum,ji,fi,t,n;
while(scanf("%f%d",&n,&m)&&n+1||m+1){
k=3*m,sum=0;
float a[k];
for(i=0;i

杭电 ACM
小弟想知道为什么循环周期是49 呢,
A number sequence is defined as follows:
f(1) = 1,f(2) = 1,f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A,B,and n,you are to calculate the value of f(n).
Input
The input consists of multiple test cases.Each test case contains 3 integers A,B and n on a single line (1

杭电let the ballon rise哪里错了
Problem Description
Contest time again!How excited it is to see balloons floating around.But to tell you a secret,the judges' favorite time is guessing the most popular problem.When the contest is over,they will count the balloons of each color and find the result.
This year,they decide to leave this lovely job to you.
Input
Input contains multiple test cases.Each test case starts with a number N (0 < N

麻烦各位帮我看一下杭电acm2034为什么总是Wrong Answer?
/*
Problem Description
参加过上个月月赛的同学一定还记得其中的一个最简单的题目,就是{A}+{B},
那个题目求的是两个集合的并集,今天我们这个A-B求的是两个集合的差,就是做集合的减法运算.
（当然,大家都知道集合的定义,就是同一个集合中不会有两个相同的元素,这里还是提醒大家一下）
Input
每组输入数据占1行,每行数据的开始是2个整数n(0

杭电acm2034 这样写哪里错了、、、急啊···
Problem Description
参加过上个月月赛的同学一定还记得其中的一个最简单的题目,就是{A}+{B},那个题目求的是两个集合的并集,今天我们这个A-B求的是两个集合的差,就是做集合的减法运算.（当然,大家都知道集合的定义,就是同一个集合中不会有两个相同的元素,这里还是提醒大家一下）
Input
每组输入数据占1行,每行数据的开始是2个整数n(0

杭电2067,题意要怎么理解?不需要你粘贴代码,只需要问题的理解.
杭电2067,题意要怎么理解,是不是指不过对角线,而到达终点的最短路径数?好像如果终点为（2,2）,那么这样的路径数为4个?为什么会是4个,请指教.
如果终点为（3,3），那么对角线下是哪五条路径？一直不理解这个。可是好像还没有人回答出我想要的，你们说的我都懂，可是这道题目你们并没有讲出精髓所在啊，希望有人能够说得明白一点，

杭电ACM 2011 素数判定 输入任意两位数,都输出了"OK"
Problem Description
对于表达式n^2+n+41,当n在（x,y）范围内取整数值时（包括x,y）(-39<=x


Input
输入数据有多组,每组占一行,由两个整数x,y组成,当x=0,y=0时,表示输入结束,该行不做处理.



Output
对于每个给定范围内的取值,如果表达式的值都为素数,则输出"OK",否则请输出“Sorry”,每组输出占一行.




Sample Input
0 1
0 0




Sample Output
OK



#include
#include
int main()
{
int m,n,a,num,i,j;
num=0;
while(scanf("%d %d",&m,&n)!=EOF)
{
if(m==0&&n==0)
break;
for(i=m;i<=n;i++)
{
num=i*i+i+41;

for(j=1;j<=num;j++)
{
if(num%j==0)
a++;
else
;
}
}
if(a!=0)
printf("OKn");
else
printf("Sorryn");
}
return 0;
}




我觉得可能是循环出了问题,

杭电acm1002解题思路.用C++
Problem Description
I have a very simple problem for you.Given two integers A and B,your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1

杭电ACM 第1009题 FatMouse' Trade
Problem Description
FatMouse prepared M pounds of cat food,ready to trade with the cats guarding the warehouse containing his favorite food,JavaBean.
The warehouse has N rooms.The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food.FatMouse does not have to trade for all the JavaBeans in the room,instead,he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food.Here a is a real number.Now he is assigning this homework to you:tell him the maximum amount of JavaBeans he can obtain.
Input
The input consists of multiple test cases.Each test case begins with a line containing two non-negative integers M and N.Then N lines follow,each contains two non-negative integers J[i] and F[i] respectively.The last test case is followed by two -1's.All integers are not greater than 1000.
Output
For each test case,print in a single line a real number accurate up to 3 decimal places,which is the maximum amount of JavaBeans that FatMouse can obtain.
Sample Input
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
Sample Output
13.333
31.500
我的代码如下：
#include
#define M 100
void main()
{
x05int m,n,i,k,j[M],f[M],t1,t2;
x05double a[M],temp,s;
x05while(scanf("%d%d",&m,&n)&&(m!=-1||n!=-1))
x05{
x05x05i=0;
x05x05k=n;
x05x05s=0;
x05x05while(k--)
x05x05{
x05x05x05scanf("%d%d",&j[i],&f[i]);
x05x05x05a[i]=(double)j[i]/f[i];
x05x05x05i++;
x05x05}
x05x05for(i=0;i

杭电acm题,怎么写
input
The inputs start with a line containing a single integer n.Each of the n following lines contains one test case.Each test case consists of three integers 1

杭电1097题目：下面是我的编程,求指教,为什么老师说time limited
Problem Description
lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius:gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that:gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.
Input
There are mutiple test cases.Each test cases consists of two numbers a and b(0

杭电ACM1003题目意思
Problem Description
Given a sequence a[1],a[2],a[3].a[n],your job is to calculate the max sum of a sub-sequence.For example,given (6,-1,5,4,-7),the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1

杭电acm2028 哪里错啦
Problem Description
求n个数的最小公倍数.
Input
输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数.
Output
为每组测试数据输出它们的最小公倍数,每个测试实例的输出占一行.你可以假设最后的输出是一个32位的整数.
Sample Input
2 4 6
3 2 5 7
Sample Output
12
70
Author
lcy
#include
long int divisor(long int m,long int n)
x05{
x05long int temp;
x05x05if(m

杭电ACM比赛的一道题目,
A.Shooshuns and Sequence
Time Limit :4000/2000ms (Java/Other) Memory Limit :524288/262144K (Java/Other)
Total Submission(s) :130 Accepted Submission(s) :34
Problem Description
One day shooshuns found a sequence of n integers,written on a blackboard.The shooshuns can perform one operation with it,the operation consists of two steps:
1 Find the number that goes k-th in the current sequence and add the same number to the end of the sequence;
2 Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input
The first line contains two space-separated integers n and k (1≤k≤n≤105).
The second line contains n space-separated integers:a1,a2,...,an (1≤ai≤105) the sequence that the shooshuns found.
Output
Print the minimum number of operations,required for all numbers on the blackboard to become the same.If it is impossible to achieve,print -1.
Sample Input
3 2
3 1 1
3 1
3 1 1
Sample Output
1
-1
Source
Codeforces
我想知道怎么去确定一个最大的操作次数,若大于这个次数就判断为一直不相等,然后输出-1?

请教杭电ACM1003的题目意思?
我看不懂题目的意思
Max Sum
Time Limit:2000/1000 MS (Java/Others) Memory Limit:65536/32768 K (Java/Others)
Total Submission(s):53856 Accepted Submission(s):12109
Problem Description
Given a sequence a[1],a[2],a[3].a[n],your job is to calculate the max sum of a sub-sequence.For example,given (6,-1,5,4,-7),the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1

杭电ACM MAX SUM题目看不懂
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38673 Accepted Submission(s): 8367
Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1:
14 1 4
Case 2:
7 1 6
就是那个举例怎么得到的搞不懂。跪求思路。好的话加分。（怕分被吞掉就不弄了）

杭电ACM 1099题 题意是什么啊?看不懂.
晕,我说的就是那道题.
但是它的英文描述我看不懂……

杭电OJ的第2007题出错了吗?求高手看看输出结果里面的 4 28 20 152都是怎么出来的?
Problem Description给定一段连续的整数,求出他们中所有偶数的平方和以及所有奇数的立方和.Input输入数据包含多组测试实例,每组测试实例包含一行,由两个整数m和n组成.Output对于每组输入数据,输出一行,应包括两个整数x和y,分别表示该段连续的整数中所有偶数的平方和以及所有奇数的立方和.你可以认为32位整数足以保存结果.Sample Input1 3 2 5Sample Output4 28 20 152

初一应用题.请列式解答.应用题
1、某商品的进价是500元,标价为725元,杭电要求一下利润不低于16%的售价打折出售,则售货员最低可以打几折出售此商品?
2、某书店在促销活动中,推出一种优惠卡,每张卡售价20元,凭此卡可享受八折优惠.有一次,他先买卡后付款,结果节省了人民币12元,那么李明同学此次购书的总价值是多少钱?
3.一队旅客乘坐汽车,要求每辆汽车的旅客人数相等,起初每辆汽车乘了22人,结果剩下1人为上车；如果有一辆汽车空着开走,那么所有旅客正好能平均分乘到其他各车上,已知每辆汽车最多只能容纳32人,求起初有多少辆汽车?有多少名旅客?

杭电1007 题干看不懂 用上面的两组数字 怎么算出下面的两组 请懂的人赐教
Problem Description
给定一段连续的整数,求出他们中所有偶数的平方和以及所有奇数的立方和.
Input
输入数据包含多组测试实例,每组测试实例包含一行,由两个整数m和n组成.
Output
对于每组输入数据,输出一行,应包括两个整数x和y,分别表示该段连续的整数中所有偶数的平方和以及所有奇数的立方和.
你可以认为32位整数足以保存结果.
Sample Input
1 3
2 5
Sample Output
4 28
20 152

杭电acm上的2062求解答
Problem Description
Consider the aggregate An= { 1, 2, …, n }. For example, A1={1}, A3={1,2,3}. A subset sequence is defined as a array of a non-empty subset. Sort all the subset sequece of An in lexicography order. Your task is to find the m-th one.

Input
The input contains several test cases. Each test case consists of two numbers n and m ( 0< n

杭电acm 1097T 这个代码为什么错误?
Input
There are mutiple test cases. Each test cases consists of two numbers a and b(0

请问杭电ACM1024题根据那个状态转移方程怎么确定初始值,以及好像根据那个转移方程每个数都要用上……为啥
就是每个数都是在那两种情况中选一种被加在里面……然后得出一个二维数组某个最大值……我对动态规划一直都不是很明白,更不要说拿它来做题,
Max Sum Plus Plus
Problem Description
Given a consecutive number sequence S1,S2,S3,S4 ...Sx,...Sn (1 ≤ x ≤ n ≤ 1,000,000,-32768 ≤ Sx ≤ 32767).We define a function sum(i,j) = Si + ...+ Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0),your task is to find m pairs of i and j which make sum(i1,j1) + sum(i2,j2) + sum(i3,j3) + ...+ sum(im,jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).
But I`m lazy,I don't want to write a special-judge module,so you don't have to output m pairs of i and j,just output the maximal summation of sum(ix,jx)(1 ≤ x ≤ m) instead.
Input
Each test case will begin with two integers m and n,followed by n integers S1,S2,S3 ...Sn.
Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
状态转移方程:B[i][j]=max{B[i][j-1]+A[j],B[i-1][t]+A[j] (i-1

杭电1164不知道我的程序为什么这样?希望大家帮我看看!
/*
Problem Description
Eddy's interest is very extensive,recently he is interested in prime number.
Eddy discover the all number owned can be divided into the multiply of prime number,
but he can't write program,so Eddy has to ask intelligent you to help him,
he asks you to write a program which can do the number
to divided into the multiply of prime number factor .
Input
The input will contain a number 1 < x<= 65535 per line
representing the number of elements of the set.
Output
You have to print a line in the output for each entry with the answer
to the previous question.
Sample Input
11
9412
Sample Output
11
2*2*13*181
*/
#include
#define MAX 5000
int n;
int boolean(int nn,int k,int *prime)
{
x05int i,flag=0;
x05for(i=0;ix05{
x05x05if(nn==prime[i])
x05x05{
x05x05x05flag=1;
x05x05x05break;
x05x05}
x05}
x05return flag;
x05
}
void calculate(int *prime,int *key,int k)
{
x05
x05int i,j;
x05for(i=0,j=0;ix05{
x05x05if(n%prime[i]==0)
x05x05{
x05x05x05key[j]=prime[i];
x05x05x05printf("%d*",key[j]);
x05x05x05j++;
x05x05x05n=n/prime[i];
x05x05x05if(boolean(n,k,prime))
x05x05x05{
x05x05x05x05
x05x05x05x05key[j]=n;
x05x05x05x05printf("%dn",key[j]);
x05x05x05x05break;
x05x05x05}
x05x05x05else
x05x05x05{
x05x05x05x05calculate(prime,key+j,k);
x05x05x05}
x05x05x05
x05x05}
x05}
}
int main(int argn,char *args[])
{


x05int i;
x05int j;
x05int k;
x05int prime[MAX];
x05int key[MAX];
x05while(scanf("%d",&n)!=EOF)
x05{
x05x05for(i=2,k=0;i<=n;i++)//用prime数组存放所有n(包括n)之前的所有素数
x05x05{
x05x05x05for(j=2;jx05x05x05{
x05x05x05x05if(i%j==0)
x05x05x05x05{
x05x05x05x05x05break;
x05x05x05x05}
x05x05x05}
x05x05x05if(j==i)
x05x05x05{
x05x05x05x05prime[k]=i;
x05x05x05x05k++;
x05x05x05}
x05x05}


x05x05if(boolean(n,k,prime))
x05x05{
x05x05x05printf("%dn",n);
x05x05}
x05x05else
x05x05{
x05x05x05calculate(prime,key,k);
x05x05x05x05
x05x05}


x05x05


x05}




x05return 0;
}




结果为什么会多一个呀?

杭电ACM,1002题,总是格式错误
Input
Input contains an integer N in the first line,and then N lines follow.Each line starts with a integer M,and then M integers follow in the same line.
我的程序：
Output
For each group of input integers you should output their sum in one line,and you must note that there is a blank line between outputs.
#include"stdio.h"
#define N 100
int main()
{int n,m,i,j,sum[N]={0},a;
scanf("%d",&m);
for(j=0;j

杭电的1005题 我不需要代码 请各位给我讲讲思路和做法
A number sequence is defined as follows:
f(1) =
1,f(2) = 1,f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A,B,and
n,you are to calculate the value of f(n).
如果就按它那么写的话会超时
我知道有很多解题报告 但是我想请你们先给我讲讲思路 然后我再试着自己去尝试

杭电OJ 3177题老是WA,
/*
思路：贪心算法,我们每次搬运想到就是使剩余的洞穴空间尽量大,当有两件物品时,有如下信息
A a1 b1
B a2 b2
加入先搬A,则搬运过程中瞬间占用空间最大为a1+b2,如果先搬运B,
则这一个瞬间最大占用体积为a2+b1,我们想使剩下的体积尽量大,所以我们选择min(a1+b2,a2+b1),如果先选A,
则有a1+b2b2-a2,所以我们可以看出,应该先选搬运体积和本身体积差值最大的贪心,由此得到最优搬运顺序!
*/
#include
typedef struct node{
int a;
int b;
int c;
}Node;
int quicksort(Node a[],int s,int e);
int main()
{
int v,n;
int t;
int i;
int flag;
Node a[1000];
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&v,&n);
for(i=0;iv)flag=0;
else v-=a[i].a;
}
//开始逐个放进洞里.
if(flag)printf("Yesn");
else printf("Non");
}
return 0;
}
int quicksort(Node a[],int s,int e)
{
int f,b,key;
Node m,n;
f=s+1;
b=e;
key=a[s].c;
n=a[s];
if(s>=e)return 0;
while(f

ACM题目：请问杭电1358的题目意思是什么?
以下是题目：
Description
For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126,inclusive),we want to know whether the prefix is a periodic string.That is,for each i (2

杭电acm2034基础题 C++,求改错
Problem Description
参加过上个月月赛的同学一定还记得其中的一个最简单的题目,就是{A}+{B},那个题目求的是两个集合的并集,今天我们这个A-B求的是两个集合的差,就是做集合的减法运算.（当然,大家都知道集合的定义,就是同一个集合中不会有两个相同的元素,这里还是提醒大家一下）
Input
每组输入数据占1行,每行数据的开始是2个整数n(0

杭电ACM 2026
输入一个英文句子,将每个单词的第一个字母改成大写字母.
Input
输入数据包含多个测试实例,每个测试实例是一个长度不超过100的英文句子,占一行.
Output
请输出按照要求改写后的英文句子.
example Input
i like acm
i want to get an acceptedSample Output
I Like Acm
I Want To Get An Accepted
代码
#include
main()
{
char a[100];
int b,c,d,i,j;
while(scanf("%s",&a)!=EOF)
{
printf("%c",a[0]-32);
for(i=1;i

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原题：http://acm.hdu.edu.cn/showproblem.php?pid=1216
assistance required
problem description
after the 1997/1998 southwestern european regional contest (which was held in ulm) a large contest *** took place. the organization team invented a special mode of choosing those participants that were to assist with washing the dirty dishes. the contestants would line up in a queue, one behind the other. each contestant got a number starting with 2 for the first one, 3 for the second one, 4 for the third one, and so on, consecutively.
the first contestant in the queue was asked for his number (which was 2). he was freed from the washing up and could *** on, but every second contestant behind him had to go to the kitchen (those with numbers 4, 6, 8, etc). then the next contestant in the remaining queue had to tell his number. he answered 3 and was freed from assisting, but every third contestant behind him was to help (those with numbers 9, 15, 21, etc). the next in the remaining queue had number 5 and was free, but every fifth contestant behind him was selected (those with numbers 19, 35, 49, etc). the next had number 7 and was free, but every seventh behind him had to assist, and so on.
let us call the number of a contestant who does not need to assist with washing up a lucky number. continuing the selection scheme, the lucky numbers are the ordered sequence 2, 3, 5, 7, 11, 13, 17, etc. find out the lucky numbers to be prepared for the next contest ***.

input
the input contains several test cases. each test case consists of an integer n. you may assume that 1

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转动惯量的测定实验,如果转轴偏离长细棒的中心,实验测得的细棒转动惯量与理论值1/2 m l方 比较是偏大还是偏小 为什么 分析讨论题 第二题

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那总分是多少啊？