C compiler Turbo C.
click here to download
Monday, December 28, 2009
Sunday, December 27, 2009
Java Compiler (jdk1.5)
u can download the java compiler jdk1.5 from the link given below. please try again later if it shows "the file is temporarily unavailable".
da link must work.
click here to download
da link must work.
click here to download
Java Compiler (jdk1.5)
u can download the java compiler jdk1.5 from the link given below. please try again later if it shows "the file is temporarily unavailable".
da link must work.
click here to download
da link must work.
click here to download
Monday, December 21, 2009
Programming Game All by Example-Mat Buckland
Programming Game All by Example by Mat Buckland. Might be useful for many.
click here to download (rapidshare link)
or
click here to download (megaupload link)
click here to download (rapidshare link)
or
click here to download (megaupload link)
Tuesday, November 10, 2009
Thursday, October 22, 2009
UVA Problem#11530
Question:http://uva.onlinejudge.org/external/115/11530.html
Solution:int main(){..........
if(s[j]=='a' || s[j]=='d' || s[j]=='g' || s[j]=='j' || s[j]=='m' || s[j]=='p' || s[j]=='t' || s[j]=='w' || s[j]==' '){
c += 1;
}else if(s[j]=='b' || s[j]=='e' || s[j]=='h' || s[j]=='k' || s[j]=='n' || s[j]=='q' || s[j]=='u' || s[j]=='x'){
c += 2;
}else if(s[j]=='c' || s[j]=='f' || s[j]=='i' || s[j]=='l' || s[j]=='o' || s[j]=='r' || s[j]=='v' || s[j]=='y'){
c += 3;
}else{
c += 4;
}........ }
UVA Problem#10222
Question:
Solutions :
#include<stdio.h>
#include<string.h>
void pri(char c){
if(c == ']'){
printf("p");
}else if(c == '['){
printf("o");
}else if(c == 'p' || c == 'P'){
printf("i");
}else if(c == 'o' || c == 'O'){
printf("u");
}else if(c == 'i' || c == 'I'){
printf("y");
}else if(c == 'u' || c == 'U'){
printf("t");
}else if(c == 'y' || c == 'Y'){
printf("r");
}else if(c == 't' || c == 'T'){
printf("e");
}else if(c == 'r' || c == 'R'){
printf("w");
}else if(c == 'e' || c == 'E'){
printf("q");
}else if(c == '\''){
printf("l");
}else if(c == ';'){
printf("k");
}else if(c == 'l' || c == 'L'){
printf("j");
}else if(c == 'k' || c == 'K'){
printf("h");
}else if(c == 'j' || c == 'J'){
printf("g");
}else if(c == 'h' || c == 'H'){
printf("f");
}else if(c == 'g' || c == 'G'){
printf("d");
}else if(c == 'f' || c == 'F'){
printf("s");
}else if(c == 'd' || c == 'D'){
printf("a");
}else if(c == '.'){
printf("m");
}else if(c == ','){
printf("n");
}else if(c == 'm' || c == 'M'){
printf("b");
}else if(c == 'n' || c == 'N'){
printf("v");
}else if(c == 'b' || c == 'B'){
printf("c");
}else if(c == 'v' || c == 'V'){
printf("x");
}else if(c == 'c' || c == 'C'){
printf("z");
}else if(c == 's' || c == 'S'){
printf("]");
}else if(c == 'a' || c == 'A'){
printf("[");
}else if(c == 'x' || c == 'X'){
printf("\\");
}else if(c == 'z' || c == 'Z'){
printf("\'");
}
}
UVA Problem#10018
Question:
Solutions :
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s1[100], s[100], s2[100];
void add(char f[], char s[], char r[]){
int lf, ls, d=0, i, j=0, c=0, sum;
lf=strlen(f); ls=strlen(s);
reverse(f, f+lf); reverse(s, s+ls);
if(lf>=ls){
d=lf-ls;
for(i=0; i<ls; i++){
sum = f[i]+s[i]-96+c;
r[j++] = (sum%10)+48;
c = sum/10;
}
for(i=ls; i<lf; i++){
sum = f[i]-48+c;
r[j++] = (sum%10)+48;
c = sum/10;
}
if(c!=0){
sum = c%10;
r[j++] = sum+48;
if(c>9)r[j++] = (c/10)+48;
}
r[j] = '\0';
lf = strlen(r);
reverse(r, r+lf);
}else{
d=lf-ls;
for(i=0; i<lf; i++){
sum = f[i]+s[i]-96+c;
r[j++] = (sum%10)+48;
c = sum/10;
}
for(i=lf; i<ls; i++){
sum = s[i]-48+c;
r[j++] = (sum%10)+48;
c = sum/10;
}
if(c!=0){
sum = c%10;
r[j++] = sum+48;
if(c>9)r[j++] = (c/10)+48;
}
r[j] = '\0';
lf = strlen(r);
reverse(r, r+lf);
}
}
UVA Problem#543
Question:
Solutions :
#include<stdio.h>
#include<math.h>
int prime(int n){
int i, c=1, j=int(sqrt(n));
for(i=2; i<=j; i++){
if(n%i==0){
c = 0;
break;
}
}
return c;
}
UVA Problem#458
Question:
Solutions :
#include<stdio.h>
#include<string.h>
char s[1000000];
int main(){
int i, l, c;
while(gets(s) != NULL){
l = strlen(s);
for(i=0; i<l; i++){
c = s[i]-7;
printf("%c", c);
}
printf("\n");
}
return 0;
}
UVA Problem#424
Question:
Solution:
char f[100000], s[100000], r[100000];
void add(char f[], char s[], char r[]){
int lf, ls, d, i, j=0, c=0, sum; lf=strlen(f);
ls=strlen(s);
reverse(f, f+lf);
reverse(s, s+ls);
UVA Problem#382
Question:
Solution:
void perfection(int n){
int i , sum=0, m=n/2;
for(i=1; i<=m; i++){
if(n%i==0)
sum += i;
}
if(n>sum){
printf("DEFICIENT\n");
}
else if(n<sum){
printf("ABUNDANT\n");
}
else if(n == sum){
printf("PERFECT\n");
}
}
int main(){
....
....
}
Friday, October 9, 2009
The Art of Computer Programming by Donald E. Knuth (Boxed_Set_Vol_1-4.part2)
The Art of Computer Programming by Donald E. Knuth. The link to download the 2nd part of the book is given below.
click here to download
click here to download
The Art of Computer Programming by Donald E. Knuth (Boxed_Set_Vol_1-4.part2)
The Art of Computer Programming by Donald E. Knuth. The link to download the 2nd part of the book is given below.
click here to download
click here to download
The Art of Computer Programming by Donald E. Knuth (Boxed_Set_Vol_1-4.part1)
The Art of Computer Programming by Donald E. Knuth. Here is the link to download the 1st part.
click here to download
click here to download
A Computational Introduction To Number Theory And Algebra
A Computational Introduction to Number Theory by Victor Shoup
click here to download
Password to unzip/unrar: www.ebooksdb.com
click here to download
Password to unzip/unrar: www.ebooksdb.com
Thursday, October 8, 2009
UVA Problem#10432
Question:
Solution technique:
1. take inputs
2. define pi as acos(-1.0)
3. use the formula below to find the output:
area = (n*r*r*(sin(2.0*pi/n)))/2;
UVA Problem#10370
Question:
Solution:
1. take inputs c, n, and the numbers in an array a[]
2. add all the elements of the array and find the average.
3. now check for the following condition and print the result:
int main(){
....
.....
for(j=0; j>n; j++){
if(a[j]<avg)count++;
}
r = (count/n)*100;
printf("%.3lf%c\n", r, d);
}Wednesday, October 7, 2009
Big Mod (uva#374)
Question:
Solutions :
long bigmod(long b, long p, long m){
if(p==0)
return 1;
else if(p%2==0)
return ((bigmod(b, (p/2), m))*(bigmod(b, (p/2), m)))%m;
else return((b%m)*bigmod(b, p-1, m))%m;
} Fibonacci using quick matrix
Solutions :
int fibo(int n){ // using quick method matrix program
int i, h, j, k, t;
i=h=1;
j=k=0;
while(n>0){
if(n%2==1){
t=j*h;
j=i*h + j*k +t;
i= i*k + t;
}
t= h*h;
h=2*h*k + t;
k= k*k +t;
n= (int) n/2;
}
return j;
}
Fibonacci using dynamic program
Solutions :
int fibo(int n){ // // using dynamic program
int a=1, b=1, i, c;
for(i=2; i>=n; i++){
c=a+b;
a=b;
b=c;
}
return a;
}
Fibonacci using normal recursion
Solutions :
int fibo(int n){ // using normal recursion
if(n<2) return (fibo(n-1)+fibo(n-2));
if(n==0) return 0;
else return (1);
}
Ins and outs of Fibonacci numbers
here is a describe document on fibonacci number. Almost everything of it....
click here to view/download
click here to view/download
Ins and outs of Fibonacci numbers
here is a describe document on fibonacci number. Almost everything of it....
click here to view/download
click here to view/download
Bit wise operator
Using of Bit wise operators instead of any other mathematical operators is much more faster. here is a document of its uses.
click here to download
click here to download
Bit wise operator
Using of Bit wise operators instead of any other mathematical operators is much more faster. here is a document of its uses.
click here to download
click here to download
Tuesday, October 6, 2009
Monday, October 5, 2009
Programming Challenges- the Programming Contest Training Manual
Programming Challenges- the Programming Contest Training Manual by Steven S.Skiena and Miguel A.Revilla.
click here to download
click here to download
Java 2-The Complete Reference (fifth edition)-Herbert Schildt
Java 2-The Complete Reference (fifth edition)-Herbert Schildt........another useful book to learn JAVA language.
click here to download
click here to download
UVA Problem#100
Question:
Solution:
1. take input i, j. let x=i, y=j.
2. Generate the series.
for(a=i; a<=j; a++){
n=a;
count=1;
while(n!=1){
if(n%2==0){
n = n/2;
}
else{
n = 3*n+1;
}
count++;
}
3. find the max count value by
if(maxcount < count){
maxcount=count;
}
JAVA How to Program (7th Edition) by Deitel and Deitel
A complete java book full of examples. "JAVA How to Program (7th Edition) by Deitel and Deitel".
click here to download
click here to download
Sunday, September 20, 2009
Saturday, September 19, 2009
UVA Problem#10812
Question:
http://uva.onlinejudge.org/external/108/10812.html
Solution:
1. run the loop down to n
2. take the input s and d.
3. Print the output using this technique:
if(s<d || (s+d)%2 || (s-d)%2)
printf("impossible\n");
else{
a=(s+d)/2;
b=(s-a);
if(a>b)
printf("%d %d\n", a, b);
else
printf("%d %d\n", b, a);
}
http://uva.onlinejudge.org/external/108/10812.html
Solution:
1. run the loop down to n
2. take the input s and d.
3. Print the output using this technique:
printf("impossible\n");
else{
a=(s+d)/2;
b=(s-a);
if(a>b)
printf("%d %d\n", a, b);
else
printf("%d %d\n", b, a);
}
Thursday, September 17, 2009
UVA Problem#11185
Question:
Solution Techniques:
1. take input.
2. use the following formula to get the ternary of the number
ternary[i] = decimal%3;
decimal = decimal/3;
i++;
3. Show the output.
UVA Problem#10071
#include <stdio.h>
int main(){
int v, t;
while(scanf("%d%d", &v, &t)==2)
printf("%d\n", 2*v*t);
return 0;
}
Wednesday, September 16, 2009
UVA Problem#11172
Question:
http://uva.onlinejudge.org/external/111/11172.html
Solution Method:
it is one of the most easiest problem of UVA.
1. take simple input
2. bye simple comparing you can achieve the proper output.
http://uva.onlinejudge.org/external/111/11172.html
Solution Method:
it is one of the most easiest problem of UVA.
1. take simple input
2. bye simple comparing you can achieve the proper output.
UVA Problem#11636
Question:
http://uva.onlinejudge.org/external/116/11636.html
Solution:
int main(){
...
.....
scanf("%d", &n);
if(n<0) break;
if(n>1){
j = 0;
line = 1;
for(j=1; ; j++){
line *= 2;
if(line >= n) break;
}
....
....}
http://uva.onlinejudge.org/external/116/11636.html
Solution:
int main(){
...
.....
scanf("%d", &n);
if(n<0) break;
if(n>1){
j = 0;
line = 1;
for(j=1; ; j++){
line *= 2;
if(line >= n) break;
}
....
....}
UVA Problem#10783
Question:
Solution Method:
1. loop will run from the given starting value till the gien end value
2. check for the odd numbers
3. simply add them
UVA Problem#575
s=1;
for(j=l; j>i; j--){
s*=2;
}
k+=((a[i]-48)*(s-1));
}
UVA Problem#438
Question:
Solution:
int main(){
.....
....
m1= (y2-y1)/(x2-x1);
m2= (y3-y1)/(x3-x1);
if (m2==0){
k= (x2-x3)/m1/2 + (y1+y2)/2;
h= (x1+x3)/2; }
if (m1==0){
k=(x3-x2)/m2/2 + (y1+y3)/2;
h=(x1+x2)/2;
}
else if (m1!=0&&m2!=0){
h= ((x1+x2)/m1 - (x1+x3)/m2 + y2-y3)/2/(1/m1 - 1/m2);
k= ( h-(x1+x2)/2)/(-m1) + (y1+y2)/2;
}
r= sqrt((h-x1)*(h-x1) + (k-y1)*(k-y1));
ans=2*pi*r;
......
...
}
UVA Problem#369
Question:
Solution:
double fact(int n){
int i; double f=1;
for(i=2; i<=n; i++){
f = f*i;
}
return f;
}
int main(){
....
....
c = fact(n) / ( fact(n-m) * fact(m) )
... }
Friday, September 11, 2009
Addison-Wesley_Concrete.Mathematics
An essential book for solving the program related to mathematics! Very usefull though very difficult!!
click here to download
click here to download
Wednesday, September 9, 2009
Introduction to Algorithm-Cormen
A book of algorithm by Cormen. Algorithms are must to do well in any kind of programming contest.
click here to download
click here to download
Art of Programming Contest
A wonderful book for the amateur contestant. This book is a must to know all the ABC of a programming contest.
Thursday, August 27, 2009
C Programming- A book from oxford
A Programming Book from OXFORD university! A small book but very useful for learning.
click here to download
click here to download
C Programming for the absolute Beginner
C Programming For The Absolute Beginner is a very useful book for the beginner!
click here to download
click here to download
Programming with C- Schaum Series
Programming with C- Schaum Series is amongst the mostly used book to learn 'C' language! its a very useful book!
click here to download
click here to download
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