ACM notes

1. ACM notes • Can work on the problems anytime throughout the term • Contest conflict -- The GRE subject test is Nov. 10th! Digroot “problem” int x; while (cin >> x) { cout << ( (x%9) ? (x%9) : 9 ); }

2. 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

3. Storing results in tables www.cs.hmc.edu/ACM/ “dynamic programming” Find the number of prime factors in N! (1 is not prime.) ... 2 3 4 5 6 7 8 ...

4. Other problems • Change counting input: output: 1.00 0.06 0 There are 292 ways to make $1.00 There are 2 ways to make $0.06 • Sigma series What? input: output: 3 4 87 99 -1 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

5. Other problems • Change counting input: output: 1.00 0.06 0 There are 292 ways to make $1.00 There are 2 ways to make $0.06 • Sigma series Shortest sequences from 1 to N such that each element is the sum of two previous elements. input: output: 3 4 87 99 -1 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

6. C++ STL www.dinkumware.com/htm_cpl/index.html www.sgi.com/tech/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 vector #include <vector> sort #include <algorithm> last time deque #include <deque>

7. Useful C functions int atoi(char* s); double atof(char* s); int strcasecomp(char* s1, char* s2); long strtol(char* s, NULL, int base) 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 strtol(“Charlie”, NULL, 36)== 2147483647L use man for more...

8. 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:

9. Two ACM programming skills A chance to “improve” your C/C++ … Preparation for the ACM competition ... Problem Insight and Execution ... 2 1 Anxiety! Get into the minds of the judges

10. Get into the minds of the judges Key Skill #1: mindreading 100% 0% “What cases should I handle?” spectrum

11. Key Skill #2: anxiety Anxiety!

12. Dynamic Programming Strategy: create a table of partial results & build on it. divis.cc T(n) = number of steps yet to go T(n) = T(3n+1) + 1 if n odd T(n) = T(n/2) + 1 if n even

13. Dynamic Programming Keys: create a table of partial results, articulate what each table cell means, then build it up... divis.cc j = items considered so far Table T 0 1 2 3 4 5 6 1 1 6 2 -3 0 the list 1 4 i = possible remainder the divisor 2 3 T[i][j] is 1 if i is a possible remainder using the first j items in the list.

14. Dynamic programs can be short #include <cstdio> #include <iostream> #include <vector> vector<int> v(10000); vector<bool> m(100); // old mods vector<bool> m2(100); // new mods int 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

15. 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 Best bugs from last week: getting the input in the “pea” problem: filling in the table in the “divis” problem: for (int j=1 ; j<N ; ++j) { cin >> Array[i]; } Table[i + n % k] = 1; Table[i - n % k] = 1;

16. New Problem Word Chains hertz jazz hajj zeroth doze aplomb ceded dozen envy ballistic yearn Input A list of words Output yes or no -- can these words be chained together such that the last letter of one is the first letter of the next… ?

17. Knapsack Problem objectwt.val. 1 3 8 2 2 5 3 1 1 4 2 5 Maximize loot w/ weight limit of 4. w Weight available for use Number of objects considered 0 1 2 3 4 1 V(n,w) = max value stealable w/ ‘n’ objects & ‘w’ weight 2 3 4 n V(n,w) =

18. C Output printf, fprintf, sprintf(char* s, const char* format, …) the destination the format string the values possible format strings h % -#0 12 .4 d d decimal integers u unsigned (decimal) ints o octal integers x hexadecimal integers f doubles (floats are cast) e doubles (exp. notation) g f or e, if exp < -3 or -4 c character s string n outputs # of chars written !! % two of these print a ‘%’ type size modifier minimum field width precision allowed size modifiers flags - left-justify 0 pad w/ zeros + use sign (+ or -) (space) use sign ( or -) # deviant operation h short l long (lowercase L) L long double start character

19. C Output value = 42 value = -42 %10.4d 0042 -0042 %-#12x 0x2a 0xffffffd6 value = 42 value = -42.419 %+10.4g +42 -42.42 %- 10.4g 42 -42.42 %-#10.4g 42.00 -42.42 value = “forty-two” %10.5s forty