ACM Block world Time Limit:1000ms Memory Limit:65536kb Descr

ACM的 “顺”序列 Time Limit:1000MS Memory Limit:32768K
Description:
贝贝5岁了.她从一堆数字卡片中选出了4张卡片：5、7、6、8.她摆布了一阵这些卡片后,发现它们可以排成比较顺的序列：5、6、7、8.她同样拿了另4张卡片：5、7、1、2,可是怎么也排不成“顺”的序列.原来,贝贝的所谓“顺”序列是我们所知道的等差数列!贝贝一边拿起一堆数字卡片,一边就在摆布它们,尝试着让它们“顺”起来,可总是有些“顺”,有些不“顺”.这个问题得靠你给她帮忙了,设计一个程序,能够判断对于给定的一堆数字,能“顺”还是不能“顺”.
Input:
输入中第一行为一个整数n（1≤n≤10）,描述后面一共有n组卡片,每组卡片的第一个数m（1≤m≤100）,表示后面会出现m张卡片.
Output:
针对每组卡片,判断是否能构成“顺”序列.如果能构成“顺”序列,则输出“yes”,否则就输出“no”.每个结果应分别不同行显示.
Sample Input:
2
4 5 7 6 8
8 1 7 3 2 8 12 78 3
Sample Output:
yes
no
————————————————————————————
下面是我写的代码：
#include
using namespace std;
void bb(int *p,int size) //冒泡排序
{
int i,temp,k=1;
for(int pass=1;pass>n)
{
int i,j;
cin>>m;
for(i=0;ia[i][j];
for(i=0;i

A+B Problem Time Limit:1000MS Memory Limit:1024K Description:Calculate a + b Input:The input will
For each pair of input integers a and b you should output the sum of a and b in one line,and with one line of output for each line in input.
Sample Input:
1 5
0 0
Sample Output:
6
刚刚学习ACM第一题就错了麻烦大家帮帮忙了
我的代码是这个
#include
using namespace std;
int main(int argc,char* argv[])
{
int a,b;
while(cin >> a >> b)
cout a >> b)
cout

英语翻译
Uniform Generator
Time Limit:1000MS Memory Limit:10000K
Total Submissions:4115 Accepted:1950
Description
Computer simulations often require random numbers.One way to generate pseudo-random numbers is via a function of the form
seed(x+1) = [ seed(x) + STEP ] % MOD
where "%" is the modulus operator.
Such a function will generate pseudo-random numbers (seed) between 0 and MOD-1.One problem with functions of this form is that they will always generate the same pattern over and over.In order to minimize this effect,selecting the STEP and MOD values carefully can result in a uniform distribution of all values between (and including) 0 and MOD-1.
For example,if STEP=3 and MOD=5,the function will generate the series of pseudo-random numbers 0,3,1,4,2 in a repeating cycle.In this example,all of the numbers between and including 0 and MOD-1 will be generated every MOD iterations of the function.Note that by the nature of the function to generate the same seed(x+1) every time seed(x) occurs means that if a function will generate all the numbers between 0 and MOD-1,it will generate pseudo-random numbers uniformly with every MOD iterations.
If STEP = 15 and MOD = 20,the function generates the series 0,15,10,5 (or any other repeating series if the initial seed is other than 0).This is a poor selection of STEP and MOD because no initial seed will generate all of the numbers from 0 and MOD-1.
Your program will determine if choices of STEP and MOD will generate a uniform distribution of pseudo-random numbers.
Input
Each line of input will contain a pair of integers for STEP and MOD in that order (1

ACM的 “顺”序列 Time Limit:1000MS Memory Limit:32768K
Description:
贝贝5岁了.她从一堆数字卡片中选出了4张卡片：5、7、6、8.她摆布了一阵这些卡片后,发现它们可以排成比较顺的序列：5、6、7、8.她同样拿了另4张卡片：5、7、1、2,可是怎么也排不成“顺”的序列.原来,贝贝的所谓“顺”序列是我们所知道的等差数列!贝贝一边拿起一堆数字卡片,一边就在摆布它们,尝试着让它们“顺”起来,可总是有些“顺”,有些不“顺”.这个问题得靠你给她帮忙了,设计一个程序,能够判断对于给定的一堆数字,能“顺”还是不能“顺”.
Input:
输入中第一行为一个整数n（1≤n≤10）,描述后面一共有n组卡片,每组卡片的第一个数m（1≤m≤100）,表示后面会出现m张卡片.
Output:
针对每组卡片,判断是否能构成“顺”序列.如果能构成“顺”序列,则输出“yes”,否则就输出“no”.每个结果应分别不同行显示.
Sample Input:
2
4 5 7 6 8
8 1 7 3 2 8 12 78 3
Sample Output:
yes
no
————————————————————————————
下面是我写的代码：
#include
using namespace std;
void bb(int *p,int size) //冒泡排序
{
int i,temp,k=1;
for(int pass=1;pass>n)
{
int i,j;
cin>>m;
for(i=0;ia[i][j];
for(i=0;i

新手请教ACM水题
Q - 数据的交换输出
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
输入n(n>n)
x09{
x09x09if(n==0)continue;
x09x09for(i=1;i>m;
x09x09x09x[i-1]=m;
x09x09x09if(i==1)min=x[0];
x09x09x09if(x[i-1]

这道题求oj题超时,求简单算法
划分方法
Time Limit:1000MS Memory Limit:65536K
Total Submit:105 Accepted:30
Description
我们知道: 15=7+8=4+5+6=1+2+3+4+5 , 这里有3中分割方法, 记 dw(15)=3.
把n分成一些连续的正整数和的方法数称为dw(n)
Input
多组测试数据,每组一个int型的n（n>0）
Output
输出dw(n)
Sample Input
13
15
Sample Output
1
3
Source
军哥
我写的代码：
#include
using namespace std;
int fun(int n)
{
int sum=0,i,j;
for(i=1;in)
cout

ACM水题,WA了,
请问错在哪里了?
对称文
Time Limit:1000MS Memory Limit:32768K
Description:
所谓对称文是字符串以中心为对称.例如,madam是对称文.字串中的字符是由数字、标点符号、空格以及英文字符（包括大小写）组成.每个字串占一行.英文字符不区分大小写,也就是说,’A’与’a’视同相等.标点符号中“{}”对称.除此之外,当字串为“000000”时,输入结束.
Sample Input:
ling 121 gnil
kkghkkhg
)aba(
000000
Sample Output:
Symmetry
Symmetry
Not symmetry
Not symmetry
# include
# include
int main(void)
{
x05char s[1000];
x05int i,j,n;
x05while(gets(s))
x05{
x05x05if(strcmp(s,"000000")==0)
x05x05x05break;
x05x05for(i=0;s[i]!=' ';i++)
x05x05x05if(s[i]>='A'&&s[i]

C++编程求和
Problem D:求和
Time Limit:1000MS Memory Limit:65536K
Total Submit:38 Accepted:5
Description
简单的a+b问题,对我们而言已不在话下,现在尝试尝试新型求和吧：
给定一个正整数,求这个整数的各个位数数值之和.
Input
输入有多组测试数据,每组测试数据包含一个正整数n（1

一道acm题,
Delete Number
Time Limit:1000MS Memory Limit:65536K
Total Submit:701 Accepted:129
Description
Given 2 integer number n and m.You can delete m digits from the number n,then number n changes to a new number n1.Tell me how to delete the number,you can get the smallest one.
For example,
m:1 n:1456
n1 may be 145,156,146,456
the smallest one is 145.Then n1 should be 145.
Input
The input consists of T test cases.The number of them (T) is given on the first line of the input file.Each test case consists of one single line containing two integer number m(1

pascal初学者提问
[j9]最经济型包装箱型号
Time Limit:1000MS Memory Limit:65536K
Total Submit:32 Accepted:23
Description
已知有A,B,C,D,E五种包装箱,为了不浪费材料,小于10公斤的用A型,大于等于10公斤小于20公斤的用B型,大于等于20公斤小于40公斤的用C型,大于等于40公斤的小于50公斤的用D型,大于等于50公斤小于80公斤的用E型.现在输入一货物的重量（小于80公斤）,找出最经济型的包装箱型号.
Input
输入只有一行,包括一个整数.
Output
输出只有一行（这意味着末尾有一个回车符号）,包括1个字符.
Sample Input
8
Sample Output
A
Source

请高手翻译一下
Financial Management
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 42448 Accepted: 20746
Description
Larry graduated this year and finally has a job. He's making a lot of money, but somehow never seems to have enough. Larry has decided that he needs to grab hold of his financial portfolio and solve his financing problems. The first step is to figure out what's been going on with his money. Larry has his bank account statements and wants to see how much money he has. Help Larry by writing a program to take his closing balance from each of the past twelve months and calculate his average account balance.
Input
The input will be twelve lines. Each line will contain the closing balance of his bank account for a particular month. Each number will be positive and displayed to the penny. No dollar sign will be included.
Output
The output will be a single number, the average (mean) of the closing balances for the twelve months. It will be rounded to the nearest penny, preceded immediately by a dollar sign, and followed by the end-of-line. There will be no other spaces or characters in the output.

A+B acm出现的问题
A+B Problem
Time Limit:1000MS Memory Limit:1024K
Description:
Calculate a + b
Input:
The input will consist of a series of pairs of integers a and b,separated by a space,one pair of integers per line,0 0 means the end of the input,and do not need to output.
Output:
For each pair of input integers a and b you should output the sum of a and b in one line,and with one line of output for each line in input.
Sample Input:
1 5
0 0
Sample Output:
6
刚刚学习ACM第一题就错了麻烦大家帮帮忙了
我的代码是这个
#include
using namespace std;
int main(int argc,char* argv[])
{
int a,b;
while(cin >> a >> b)
cout a >> b)
cout

一道acm题,
Problem H:Groups (I)
Time Limit:1000MS Memory Limit:65536K
Total Submit:93 Accepted:31
Description
　　There are n people in the class and
person pi has his value vi.Now,we'd like to put them into different groups.In
each group,there must be two people,and the value of this group is equal to
the minimum of two member's values.For example,if pi and pj are in the same
group,the value of their group is min(pi,pj).Now,your task is to tell the
maximum and minimum of the sum of all groups.
Input
　　There are several test cases.
　　In
each case,there is an even integer n(n

计算题 Time Limit: 1000MS Memory limit: 65536K
题目描述
一个简单的计算,你需要计算f(m,n),其定义如下：
当m=1时,f(m,n)=n；
当n=1时,f(m,n)=m；
当m>1,n>1时,f(m,n)= f(m-1,n)+ f(m,n-1)
输入
第一行包含一个整数T（1

ACM题
英雄联盟
Time Limit:1000MS Memory Limit:32768K
Description:
There's a game called League of Legends.It's a game like DotA.
Now you have a simple task - Draw an icon of "League of Legends" by character.
The icon is like that:
​
You should draw a border(N * N,N is length of a side) that is width 1 and leave a space inside.Then you should draw an"L".
Input:
This problem has several cases.Input until EOF.Each case include a character and an integer N(7

你能帮我改改这个题吗?
Problem A:Counting
Time Limit:1000MS Memory Limit:65536K
Total Submit:14 Accepted:12
Description
一本书的页码从自然数1 开始顺序编码直到自然数n.书的页码按照通常的习惯编排,每个页码都不含多余的前导数字0.例如,第6 页用数字6 表示,而不是06 或006 等.数字计数问题要求对给定书的总页码n,计算出书的全部页码中分别用到多少次数字0,1,2,…,9.
编程任务：
给定表示书的总页码的10 进制整数n (1≤n≤10000) .计算书的全部页码中分别用到多少次数字0,1,2,…,9.
Input
输入由多组测试数据组成.
每组测试数据输入只有1 行,给出表示书的总页码的整数n.
Output
对应每组输入,输出共有10行,在第k行输出页码中用到数字k-1 的次数,k=1,2,…,10.
Sample Input
11
Sample Output
1
4
1
1
1
1
1
1
1
1
代码是
#include
main()
{ char str[5];
x05int n,i,j,m,a,b,c,d,e,f,g,h,k,l;
x05while(scanf("%d",&n)!=EOF){
x05x05a=b=c=d=e=f=g=h=k=l=0;
x05x05for(i=1;i