杭电OJ 1036 Average is not Fast Enough!

杭电 1020 无法AC
Problem Description
Given a string containing only 'A' - 'Z',we could encode it using the following method:
1.Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.
2.If the length of the sub-string is 1,'1' should be ignored.
Input
The first line contains an integer N (1 n;
x05while(n--)
x05{
x05x05t=0;
x05x05t1=0;
x05x05i=0;
x05x05gets(a);
x05x05while(a[i]!=NULL)
x05x05{
x05x05x05count=1;
x05x05x05while(a[i]==a[i+1])
x05x05x05{
x05x05x05x05count++;
x05x05x05x05i++;
x05x05x05}
x05x05x05
x05x05x05if(count==1)
x05x05x05x05cout

杭电acm 1019
problem description
the least common multiple (lcm) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set.for example,the lcm of 5,7 and 15 is 105.
input
input will consist of multiple problem instances.the first line of the input will contain a single integer indicating the number of problem instances.each instance will consist of a single line of the form m n1 n2 n3 ...nm where m is the number of integers in the set and n1 ...nm are the integers.all integers will be positive and lie within the range of a 32-bit integer.
output
for each problem instance,output a single line containing the corresponding lcm.all results will lie in the range of a 32-bit integer.
sample input
2
3 5 7 15
6 4 10296 936 1287 792 1
sample output
105
10296
#include
#include
#include
int ***(int a,int b)
{
x05int t=0;
x05if(a

杭电 2055 An easy problem
Problem Description
we define f(A) = 1,f(a) = -1,f(B) = 2,f(b) = -2,...f(Z) = 26,f(z) = -26;
Give you a letter x and a number y ,you should output the result of y+f(x).
Input
On the first line,contains a number T.then T lines follow,each line is a case.each case contains a letter and a number.
Output
for each case,you should the result of y+f(x) on a line.
Sample Input
6
R 1
P 2
G 3
r 1
p 2
g 3
Sample Output
19
18
10
-17
-14
-4
我的代码：这段一直过不了,检查不出错,无奈之下switch了一大串,终于通过了,不过还想求教这段到底哪里有问题,
#include
int main()
{
x05int a,s,i;
x05char z;
while(scanf("%d",&a)!=EOF)
{
x05for(i=0;i='a'&&z='A'&&z

杭电ACM1005 Number Sequence
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ACM 1005 java
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).
The input consists of multiple test cases.Each test case contains 3 integers A,B and n on a single line (1

杭电ACM1092题目 Problem Description Your task is to Calculate the sum of some integers.Input Input
#include"stdio.h"
int main()
{
int a,b,n;
a=0;
while(scanf("%d",&n)!=EOF){
while(n--){
scanf("%d",&b);
a=a+b;
}
printf("%dn",a);
}
return 0;
}
错在哪里啊啊,说的基础点的额

杭电acm
Problem Description
Now give you two integers n m,you just tell me the m-th number after radix point in 1/n,for example n=4,the first numble after point is 2,the second is 5,and all 0 followed
Input
Each line of input will contain a pair of integers for n and m(1

杭电 acm 1019!WR!
Problem Description
The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set.For example,the LCM of 5,7 and 15 is 105.
Input
Input will consist of multiple problem instances.The first line of the input will contain a single integer indicating the number of problem instances.Each instance will consist of a single line of the form m n1 n2 n3 ...nm where m is the number of integers in the set and n1 ...nm are the integers.All integers will be positive and lie within the range of a 32-bit integer.
Output
For each problem instance,output a single line containing the corresponding LCM.All results will lie in the range of a 32-bit integer.
Sample Input
2
3 5 7 15
6 4 10296 936 1287 792 1
Sample Output
105
10296
代码：
#include
void main()
{
x05long int n,m,i,a[1000],max,s,j,x;
x05while(scanf("%ld",&n)!=EOF)
x05{
x05x05while(n--)
x05x05{
x05x05x05scanf("%ld",&m);
x05x05x05for(i=0;i