poj3982,

## Problem:

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3612 Accepted: 1571
Description

Input

Output

Sample Input
`1 1 1`
Sample Output
`69087442470169316923566147`
Source

## Solution:

Use string instead of the long long. The system won't allocate that much memory for it.

## Source Code:

//Tue Nov 23 14:30:01 CST 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

{
int sz = max(A.size(), B.size());
int flag = 0;
if(sz > (int)A.size()) A += string(sz-A.size(), '0');
if(sz > (int)B.size()) B += string(sz-B.size(), '0');
string ret = "";
for(int i=0; i<sz; i++)
{
int temp = A[i]-'0' + B[i] - '0' + flag;
flag = temp > 9 ? 1 : 0;
temp %= 10;
ret += ('0' + temp);
}
if(flag == 1)
ret += ('0' + flag);
//  cout << A << ", " << B << ", " << ret << endl;
return ret;
}

int main(int argc, const char* argv[])
{
//  freopen("input.in", "r", stdin);
//  freopen("output.out", "w", stdout);
string a0, a1, a2;
while (cin >> a0 >> a1 >> a2)
{
vector<string> ret(100, "");
reverse(a0.begin(), a0.end());
reverse(a1.begin(), a1.end());
reverse(a2.begin(), a2.end());
ret[0] = a0;
ret[1] = a1;
ret[2] = a2;
for (int i = 3; i < 100; i++)
{
}
reverse(ret[99].begin(), ret[99].end());
cout << ret[99] << endl;
}
//  fclose(stdin);
//  fclose(stdout);
return 0;
}

poj3981,

## Problem:

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3931 Accepted: 1933
Description

Input

Output

Sample Input
`you are what you do`
Sample Output
`we are what we do`
Source

## Solution:

Straight forward.

## Source Code:

//Tue Nov 23 14:10:28 CST 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

int main(int argc, const char* argv[])
{
//  freopen("input.in", "r", stdin);
//  freopen("output.out", "w", stdout);
string str;
while (getline(cin, str))
{
int sz = str.size();
string ret = "";
for (int i = 0; i < sz;)
{
if (str[i] != 'y')
{
ret += str[i];
i++;
}
else
{
if (i + 2 < sz && str[i + 1] == 'o' && str[i + 2] == 'u')
{
ret += "we";
i += 3;
}
else
{
ret += str[i];
i++;
}
}
}
cout << ret << endl;
}
//  fclose(stdin);
//  fclose(stdout);
return 0;
}

poj3980,

## Problem:

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4360 Accepted: 2732
Description

Input

Output

Sample Input
```5 3
100 2```
Sample Output
```2
0```
Source

## Solution:

Straight forward.

Take care of: 1st, use scanf instead of cin.

## Source Code:

//Tue Nov 23 13:57:52 CST 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

int mod(int m, int n)
{
return m%n;
}

int main( int argc, const char* argv[] )
{
//  freopen("input.in", "r", stdin);
//  freopen("output.out", "w", stdout);
int a, b;
while(scanf("%d%d", &a, &b) != EOF)
{
printf("%d\n", mod(a, b));
}
//  fclose(stdin);
//  fclose(stdout);
return 0;
}

poj3979,

## Problem:

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5903 Accepted: 1884
Description

Input

Output

Sample Input
```1/8+3/8
1/4-1/2
1/3-1/3```
Sample Output
```1/2
-1/4
0```
Source

## Solution:

PS: Take care of each situation that could happen.

1st, if the numerator is 0;

2nd, if they have gcd greater than 1 after the operation.

## Source Code:

//Mon Nov 22 23:32:14 CST 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}

int main(int argc, const char* argv[])
{
//  freopen("input.in", "r", stdin);
//  freopen("output.out", "w", stdout);
int a, b, c, d;
char ch1, ch2;
while (cin >> a >> ch1 >> b >> ch2 >> c >> ch1 >> d)
{
//cout << a << b << c << d << endl;

int temp = b / gcd(b, d) * d;
int flag = 1;
switch (ch2)
{
case '+':
a = a * (temp / b) + c * (temp / d);
break;
case '-':
a = a * (temp / b) - c * (temp / d);
flag = a < 0 ? -1 : 1;
a = a < 0 ? -a : a;
}
int temp2 = gcd(a, temp);
//cout << temp2 << endl;
a /= temp2;
temp /= temp2;
if(a == 0)  //One WA here w/o considering if a is 0;
{
cout << "0" << endl;
continue;
}
if(temp == 1)
{
cout << a*flag << endl; //One WA here w/o *flag;
continue;
}
cout << a * flag << "/" << temp << endl;
}
//  fclose(stdin);
//  fclose(stdout);
return 0;
}

poj3507,

## Problem:

Judging Olympia
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5058 Accepted: 2430
Description
For years, a group of Regional Contest Directors (RCDs) of the ACM International Collegiate Programming Contest (ICPC) have been unsatisfied with the way contest submissions get ranked. The group sees it is academically wrong to emphasize the importance of program correctness, disregarding the “quality” of the program itself. After all, programming as a profession promotes design, style, maintainability, etc. and not just correctness. The group’s suggestion is to have a panel of six judges. Each judge is assigned the task of grading the submissions based on a particular aspect: 1) Correctness; 2) Robustness; 3) Overall design; 4) Clarity; 5) Coding style; and finally 6) Maintainability. The final grade of a submission would be the average of the six grades it gets.
The old guards of the current ICPC judging style have always responded that it is not possible to impartially judge a program on anything but correctness. How can the ICPC be certain that judging is fair? In other words, how can the ICPC be sure that non of the judges is favoring certain teams and disadvantaging others? Any hint of accusation to the judging process and ICPC loses the prestigious status it worked on for years. (Alright! So they do have a point.) Still, this hasn’t stopped other domains from judging candidates based on subjective metrics. Take for example Gymnastics, or The Nobel Prizes, or even the ACM’s very own Doctoral Dissertation Award. These are all highly respected awards where the winner is selected by judges using subjective metrics. ICPC could use a new judging system based on what is used in gymnastics. Rather than having each judge grade a certain aspect of the program, each of the six judges would assign an overall grade (out of ten) based on all of the six metrics mentioned above. To enforce impartiality, the final grade of a submission would be calculated as the average of all the grades after deleting two grades: The highest and the lowest. Any judge that favors a certain team (and assigns them an undeserved high grade,) risks the possibility of that grade being dismissed. Similarly, any judge that attempts to disadvantage a team by assigning them a low grade faces a similar risk.
Write a program to print the final grade of a submission.
Input
Your program will be tested on one or more test cases. Each test case is described on a single input line listing the grades of the judges. The end of the test cases is identified with a dummy test case with all the grades being zero.
Output
For each test case, print the grade on a separate line (without unnecessary decimal points and/or zeros.)
Sample Input
```8 8 8 4 4 4
8 8 6 4 4 3
0 0 0 0 0 0```
Sample Output
```6
5.5```
Source

## Solution:

Easy implementation;

## Source Code:

//Sun Nov 14 20:37:24 CST 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

int main( int argc, const char* argv[] )
{
//  freopen("input.in", "r", stdin);
//  freopen("output.out", "w", stdout);
int a[6];
while(cin >> a[0] >> a[1] >> a[2] >> a[3] >> a[4] >> a[5] && (a[0]||a[1]||a[2]||a[3]||a[4]||a[5]))
{
int mmax = a[0];
int mmin = a[0];
int sum = a[0];
for(int i=1; i<6; i++)
{
mmax = max(mmax, a[i]);
mmin = min(mmin, a[i]);
sum += a[i];
}
cout << 1.0*(sum - mmax - mmin)/4. << endl;
}
//  fclose(stdin);
//  fclose(stdout);
return 0;
}