ACM notes

• Can work on the problems anytime throughout the term

• Contest conflict -- The GRE subject test is Nov. 10th!

int x;

while (cin >> x) {

cout << ( (x%9) ? (x%9) : 9 );

}

Digroot “problem”

ACM “style”while (nlines-- > 0) { queue<char> q; string s; getline(cin, s); for (int i = 0; s[i]; i++) { if (!q.empty() && s[i] == q.front()) { // found q.pop(); } else { q.push(s[i]); if (q.size() > 10) { cout << "Not consistent with the theory\n"; goto done; } } } if (q.empty()) cout << "An echo string with buffer size ten\n"; else cout << "Not an echo string, but still consistent with the theory\n"; done: ;

}

Echo problem

Storing results in tables

“dynamic programming”

...

...

Find the number of prime factors in N!(1 is not prime.)

2 3 4 5 6 7 8

www.cs.hmc.edu/ACM/

Other problems

• Change counting

input: 1.00

0.06

0

output: There are 292 ways to make $1.00

There are 2 ways to make $0.06

• Sigma series

input: 3

4

87

99

-1

output: 1 2 3

1 2 4

1 2 4 8 16 24 28 29 58 87

1 2 4 8 16 32 33 66 99

What?

Other problems

• Change counting

input: 1.00

0.06

0

output: There are 292 ways to make $1.00

There are 2 ways to make $0.06

• Sigma series

input: 3

4

87

99

-1

output: 1 2 3

1 2 4

1 2 4 8 16 24 28 29 58 87

1 2 4 8 16 32 33 66 99

Shortest sequences from 1 to N such that each element is the sum of two previous elements.

C++ STL

vector<int> v; // basically an int array

v.reserve(10); // assure 10 spots

v.push_back(42); // adds 42 to the end

v.back(); // returns 42

v.pop_back(); // removes 42

v.size(); // # of elements

v[i]; // ith element

sort( v.begin(), v.end() ); // default sort

sort( v.begin(), v.end(), mycompare );

deque<int> d; // double-ended queue

d.push_front(42); // add to front

d.front(42); // return front element

d.push_front(42); // remove from front

sort#include <algorithm>

vector#include <vector>

deque#include <deque>

last time

www.dinkumware.com/htm_cpl/index.html www.sgi.com/tech/stl/

Useful C functions

int atoi(char* s);

double atof(char* s);

int strcasecomp(char* s1, char* s2);

long strtol(char* s, NULL, int base)

strtol(“Charlie”, NULL, 36) == 2147483647L

converts C strings to ints atoi(“100”) == 100

converts C strings to doubles atoi(“100.0”) == 100.0

case-insensitive C string comparison strcasecmp(“aCm”,“ACm”) == 0

arbitrary conversion from a string in bases (2-36) to a long int

use man for more...

sprintf

int sprintf(char* str, char* format, ...);prints anything to the string str

char str[100];

sprintf(str,“%d”,42); // str is “42”

sprintf(str,“%f”,42.0); // str is “42.0”

sprintf(str,“%10d”,42); // str is “ 42”

sprintf(str,“%-10d”,42); // str is “42 ”

flexible formatting:

right/left justify:

A chance to “improve” your C/C++ …

Preparation for the ACM competition ...

Problem Insight and Execution ...

Get into the minds of the judges

Anxiety!1 2

Two ACM programming skills

Get into the minds of the judges

Key Skill #1: mindreading

“What cases should I handle?” spectrum

100%0%

Key Skill #2: anxiety

Anxiety!

Dynamic Programming

Strategy: create a table of partial results & build on it.

divis.cc

T(n) = T(3n+1) + 1 if n odd

T(n) = number of steps yet to go

T(n) = T(n/2) + 1 if n even

Dynamic Programming

Keys: create a table of partial results, articulate what each table cell means, then build it up...

divis.cc0 1 2 3 4 5 6

0

1

2

T[i][j] is 1 if i is a possible remainder using the first j items in the list.

Table T

3

j = items considered so far

i = p

ossi

ble

rem

aind

er

1 1 6 2 -3the list

the divisor4

Dynamic programs can be short

#include <cstdio>#include <iostream>#include <vector>

vector<int> v(10000);vector<bool> m(100); // old modsvector<bool> m2(100); // new modsint n, k;

bool divisible(){ fill(m.begin(),m.end(),false); m[0] = true;

for (int i=0; i<n; i++) {

/* not giving away all of the code */ /* here the table is built (6 lines) */ }

return m[0];}

int main(){ cin >> n; // garbage

while (cin >> n) { cin >> k;

for (int i=0; i<n; i++) { cin >> v[i]; v[i] = abs(v[i]); v[i] %= k; }

cout << (divisible() ? "D" : "Not d") << "ivisible\n"; } cout << endl;}

acknowledgment: Matt Brubeck

STL: http://www.sgi.com/Technology/STL

General ACM Programming

Try brute force first (or at least consider it)

-- sometimes it will work fine…

-- sometimes it will take a _bit_ too long

-- sometimes it will take _way_ too long

for (int j=1 ; j<N ; ++j){ cin >> Array[i];}

Table[i + n % k] = 1;Table[i - n % k] = 1;

filling in the table in the “divis” problem:

getting the input in the “pea” problem:

Best bugs from last week:

New Problem

Input A list of words

Word Chains

Output yes or no -- can these words be chained together such that the last letter of one is the first letter of the next… ?

doze

aplomb

ceded

dozen

envy

ballistic

yearn

hertz

jazz

hajj

zeroth

Knapsack Problem

0 1 2 3 4

1

2

3

V(n,w) = max value stealable w/ ‘n’ objects & ‘w’ weight

V(n,w) =

object wt. val. 1 3 8 2 2 5 3 1 1 4 2 5

Maximize loot w/ weight limit of 4.

4

Number of objects

considered

Weight available for use

n

w

C Output

printf, fprintf, sprintf(char* s, const char* format, …)the destination the format string the values

h.412-#0% d

start character

flags- left-justify0 pad w/ zeros+ use sign (+ or -)(space) use sign ( or -)# deviant operation

minimum field width

precision

size modifier

h shortl long (lowercase L)L long double

type

d decimal integersu unsigned (decimal) intso octal integersx hexadecimal integersf doubles (floats are cast)e doubles (exp. notation)g f or e, if exp < -3 or -4c characters stringn outputs # of chars written !!% two of these print a ‘%’

allowed size modifiers

possible format strings

C Output

%10.4d

value = 42 value = -42

0042 -0042

%-#12x 0x2a 0xffffffd6

%+10.4g +42 -42.42

value = 42 value = -42.419

%- 10.4g 42 -42.42

%-#10.4g 42.00 -42.42

value = “forty-two”

%10.5s forty