[原创]引导C++初学者走进ACM - C++教室 - C++论坛

HJin

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第 11 楼

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This introduction to acm is very good.

+1

To apib2007 --- I tried to "sticky" this post but did not get it. Could you do that?

[此贴子已经被作者于2007-7-20 16:19:02编辑过]

I am working on a system which has no Chinese input. Please don\'t blame me for typing English.

2007-07-20 15:47

一番宝瓶

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第 13 楼

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回复：（HJin）four consecutive integers problemtar...

umm, smart and detailed analysis ... ,GO ON.

Time Limit:3s Memory Limit:8192K In/Out:Stdin/stdout

There is chessboard which has N * M squares size of 1 * 1 and small squares can build up to a large one. Can you tell Mr. Guo that how many squares in the chessboard?

For Example, when N = 2 and M = 3, there are 6 squares size of 1 * 1 and 2 squares size of 2 * 2 in the chessboard, so the answer is 8.
Input
There are several lines in the input, each line consist of two positive integers N and M (1 <= N <= 100, 1 <= M <= 100) except the last line which N = 0 and M = 0 implying the end of the input and you should not process it.
Output
Please output the number of the squares for each chessboard in a single line.

Sample Input
1 1
2 3
3 3
0 0

Sample Output
1
8
14

程序代码：


#include <iostream>
using namespace std;
#define max(x, y) ((x)>(y)?(x):(y))
int num(int, int);
int main()
{
 int N, M, ar[50];
 int i=0;
 while( cin>>N>>M )
 {
  if( N<0|| N>100 || M<0 || M>100 ) exit(-1);
  ar[i]=N;
  ar[i+1]=M;
  if(ar[i]==0 && ar[i+1]==0 ) break;
  i+=2;
 }
 for(int j=0; ar[j]!=0 && ar[j+1]!=0; j+=2, cout<<endl)
  cout<<num(ar[j], ar[j+1]);
 return 0;
}
int num(int n, int m)
{
int sum = 0;
for(int i=2; i<=max(n, m); ++i)
if( n >= i && m >= i)
{
 if( n*m ==i*i ) sum += 1;
 else 
 sum += ( m-i+1 )*( n-i+1 );
}
return sum+n*m;
}

Code:416B Time(s):0.00 Mem(k):812/2316

2.3.2.2.2.0

2007-07-20 20:23

leeco

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第 15 楼

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回复：（Jackzdong）就是不知道为什么别人的CODE用的内...

不要过多的关注开销，题目已经有限制，一般要求动态规划的，搜索绝对过不了，所以只要过了就说明算法复杂度要求达到了，最多差个常数而已。
你可能花1个小时让你的代码少用了0.0几秒，感觉还不如再做一道题好

2007-07-21 01:08

一番宝瓶

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第 17 楼

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Time Limit:3s Memory Limit:16384K In/Out:Stdin/stdout

A Fibonacci sequence is calculated by adding the previous two members of the sequence, with the first two members being both 1.

f(1)=1, f(2)=1, f(n>2)=f(n-1)+f(n-2)

Your task is to take a number as input, and print that fibonacci number.

Sample Input

100

Sample Output

354224848179261915075

Note:
No generated fibonacci number in excess of 1000 digits will be in the test data, i.e. f(20)=6765 has 4 digits.

#include <iostream>
#include <string>
#include <cassert>
#include <algorithm>
using namespace std;

string LongIntSum(string, string);
int main()
{
int n;
cin>>n;
string *str =new string[n];
for(int i=0; i<n; ++i)
( i < 2 ) ? str[i] = '1' : str[i] = LongIntSum( str[i-2], str[i-1] );
assert(str[n-1].size() < 1000);
cout<<str[n-1]<<endl;
delete []str;
return 0;
}

string LongIntSum(string s1, string s2)
{
register int i, j;
int sum, c=0;
string s = "";
if( s1.size() > s2.size() ) swap(s1, s2);
for(i = s1.size()-1, j = s2.size()-1; i >=0; --i, --j)
{
sum = s1[i] - '0' + s2[j] - '0' + c;
if( sum >= 10 ) {sum -= 10; c = 1;}
else c = 0;
s += sum + '0';
}

for(i =s2.size()-s1.size()-1; i >=0; --i)
{
sum = s2[i] - '0' + c;
if( sum >= 10 ) {sum -= 10; c = 1;}
else c = 0;
s += sum + '0';
}
if( c ) s+= c + '0';
reverse(s.begin(), s.end());
return s;
}

Code:959B Time(s):0.39 Mem(k):832/1044

[此贴子已经被作者于2007-7-21 15:41:13编辑过]

2.3.2.2.2.0

2007-07-21 15:34

爱上鱼的水

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第 19 楼

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怎样在那些做题目的网站里看别人的答案啊?

2007-09-15 10:49