Binary multiplication #define Sj (strlen(B)-j-1) #define Si (strlen(A)-i-1) int main() { int i, T[1000]; char A[300], B[300]; while (cin >> A >> B) { for (i=0 ; i<1000 ; ++i) T[i] = 0; for (i=0 ; A[i] != 0 ; ++i) if (A[Si] == '1') for (int j=0 ; B[j] != 0 ; ++j) if (B[Sj] == '1') ++T[999-i-j]; /* add columns */ /* then simplify and print */ } } 1001110 times 11001 create a table 0000000000001001110 0000000000000000000 0000000001001110000 0000000010011100000 0000000011013211110 add the columns 0000000011110011110 simplify STL: http://www.sgi.com/Technology/STL
when you see this, catch up with any pending operations Fraction calculator getNum(…) getOp(…) operate(n1,d1,n2,d2,op) simplify(int& n,int& d) gets a number or returns an error functions to help parse... gets an operator or returns an error does the arithmetic reduces n/d to lowest terms 18 - --5 / 7 * 3 + 1 - 3 ... evaluate as you go getNum(…) when you see this, catch up with any pending operations getOp(…) 18 - 15/7 pending immediate 111/7 + pending
allowed size modifiers C Output printf, fprintf, sprintf(char* s, const char* format, …) the destination the format string the values possible format strings % -#0 12 .4 h 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 ‘%’ minimum field width size modifier type 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
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
Stuff I’d never heard of... sprintf(s,”%*.*f”,width,prec,number) if width = 10, prec = 4, and number = 3.14159, then s == “ 3.142” What if you want to know how much was written to s ? sprintf(s,”%*.*f %n”,width,prec,number,&n) after which n == 11
Bugs & No Bugs class Line { Point* p1; Point* p2; #include <stdio.h> char *T="IeJKLMaYQCE]jbZRskc[SldU^V\\X\\|/_<[<:90!\"$434-./2>]s", K[3][1000],*F,x,A,*M[2],*J,r[4],*g,N,Y,*Q,W,*k,q,D;X(){r [r [r[3]=M[1-(x&1)][*r=W,1],2]=*Q+2,1]=x+1+Y,*g++=((((x& 7) -1)>>1)-1)?*r:r[x>>3],(++x<*r)&&X();}E(){A||X(x=0,g =J ),x=7&(*T>>A*3),J[(x[F]-W-x)^A*7]=Q[x&3]^A*(*M)[2 +( x&1)],g=J+((x[k]-W)^A*7)-A,g[1]=(*M)[*g=M[T+=A ,1 ][x&1],x&1],(A^=1)&&(E(),J+=W);}l(){E(--q&&l () );}B(){*J&&B((D=*J,Q[2]<D&&D<k[1]&&(*g++=1 ), !(D-W&&D-9&&D-10&&D-13)&&(!*r&&(*g++=0) ,* r=1)||64<D&&D<91&&(*r=0,*g++=D-63)||D >= 97&&D<123&&(*r=0,*g++=D-95)||!(D-k[ 3] )&&(*r=0,*g++=12)||D>k[3]&&D<=k[ 1] -1&&(*r=0,*g++=D-47),J++));}j( ){ putchar(A);}b(){(j(A=(*K)[D* W+ r[2]*Y+x]),++x<Y)&&b();}t () {(j((b(D=q[g],x=0),A=W) ), ++q<(*(r+1)<Y?*(r+1): Y) )&&t();}R(){(A=(t( q= 0),'\n'),j(),++r [2 ]<N)&&R();}O() {( j((r[2]=0,R( )) ),r[1]-=q) && O(g-=-q) ;} C(){( J= gets (K [1]))&&C((B(g=K[2]),*r=!(!*r&&(*g++=0)),(*r)[r]=g-K[2],g=K[2 ],r[ 1]&& O()) );;} main (){C ((l( (J=( A=0) [K], A[M] =(F= (k=( M[!A ]=(Q =T+( q=(Y =(W= 32)- (N=4 )))) +N)+ 2)+7 )+7) ),Y= N<<( *r=! -A)) class Line { Point* p1; Point* p2; Line(Point* p1, Point* p2) p1 = new Point(); p2 = new Point(); this.p1 = p1; this.p2 = p2; } // other methods 2000 IOCCC winner “Best use of flags” www.cs.hmc.edu/~dodds/anderson.c
“Power Towers” 20000 3 Input: n b Idea: 20000 3 Input: n b Idea: Write n in base b -- including all exponents! 9 5 3 2 20000 == 3 + 3 + 2*3 + 2*3 + 2 Output: 2 3 3+2 3 2 20000 = 3 +3 +2*3 +2*3 +2