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September 13, 2011 15:22
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MIPT EJudge problems
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/* | |
MIPT, task #000 | |
"A + B" | |
12.09.2011 | |
*/ | |
#include <stdio.h> | |
int main () | |
{ | |
int a, b; | |
scanf ("%d%d", &a, &b); | |
printf ("%d\n", a + b); | |
return 0; | |
} |
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/* | |
MIPT, task #001 | |
"Maximum" | |
12.09.2011 | |
*/ | |
#include <stdio.h> | |
#define MAX(a, b) (((a) > (b))? (a) : (b)) | |
int main() | |
{ | |
int max, a; | |
scanf ("%d", &max); | |
while (1 == scanf ("%d", &a)) { | |
max = MAX (max, a); | |
} | |
printf ("%d\n", max); | |
return 0; | |
} |
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/* | |
MIPT, task #002 | |
"Set Intersection" | |
12.09.2011 | |
*/ | |
#include <stdio.h> | |
#define MAX_SET_SIZE 1000000 | |
int main() | |
{ | |
char s[MAX_SET_SIZE + 1] = {0}; | |
int a; | |
while (scanf ("%d", &a) && a != -1) { | |
s[a] = 1; | |
} | |
while (scanf ("%d", &a) && a != -1) { | |
if (1 == s[a]) { | |
s[a] = 2; | |
} | |
} | |
/* | |
Now if s[a] == 1, then a belongs to the first set, | |
and if s[a] == 2, then a belongs to the second one. | |
*/ | |
char is_intersection = 0; | |
for (a = 0; a <= MAX_SET_SIZE; ++a) { | |
if (2 == s[a]) { | |
is_intersection = 1; | |
printf ("%d ", a); | |
} | |
} | |
if (!is_intersection) { | |
puts("empty"); | |
} | |
return 0; | |
} |
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; MIPT, task # 005 | |
; Random descending a tree | |
; 12.09.2011 | |
(define (f tree k) | |
(if (list? tree) | |
(+ (f (car tree) (* k 2.0)) | |
(f (cadr tree) (* k 2.0))) | |
(/ tree k))) | |
(display (f (read) 1)) |
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/* | |
MIPT, task # 006 | |
"Three Squares" | |
12.09.2011 | |
*/ | |
#include <stdio.h> | |
#include <math.h> | |
#define SQR(a) ((a) * (a)) | |
int main () | |
{ | |
int n, i; | |
scanf ("%d", &n); | |
int ans = 0; | |
while (n >= 0) { | |
char found = 0; | |
int a, b, c; | |
for (a = 0; SQR (a) <= n && !found; ++a) { | |
for (b = 0; SQR (a) + SQR (b) <= n && !found; ++b) { | |
c = n - SQR (a) - SQR(b); | |
float sqrt_c = sqrt(c); | |
if (sqrt_c == (int)(sqrt_c)) { | |
found = 1; | |
//printf ("%d + %d + %d = %d\n", SQR (a), SQR (b), SQR (c), n); | |
} | |
} | |
} | |
if (!found) { | |
++ans; | |
} | |
--n; | |
} | |
printf ("%d", ans); | |
return 0; | |
} |
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/* MIPT, task # 008 | |
"Brackets" | |
13.09.2011 | |
*/ | |
#include <stdio.h> | |
#define MAX_STR_LENGTH 4000 | |
int main() | |
{ | |
char s[MAX_STR_LENGTH]; | |
gets (s); | |
int balance = 0; | |
char *cur = s; | |
while (*cur) { | |
if ('(' == *cur) { | |
++balance; | |
} else { | |
--balance; | |
} | |
if (balance < 0) { | |
puts ("NO"); | |
exit (0); | |
} | |
++cur; | |
} | |
if (!balance) { | |
puts ("YES"); | |
} else { | |
puts ("NO"); | |
} | |
return 0; | |
} |
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; MIPT, task # 009 | |
; "Fibonacci numbers" | |
; 13.09.2011 | |
(define (fib n) | |
(define (fib-iter i a b) | |
(if (= i n) | |
a | |
(fib-iter (1+ i) b (+ a b)))) | |
(fib-iter 0 1 1)) | |
(display (fib (read))) |
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/* | |
MIPT, task # 027 | |
"Odd number" | |
13.09.2011 | |
*/ | |
#include <stdio.h> | |
#include <stdlib.h> | |
#define MAX_NUMBER 1000000 | |
char odd[MAX_NUMBER + 1]; | |
int main() | |
{ | |
int n; | |
scanf ("%d", &n); | |
while (n--) { | |
int num; | |
scanf ("%d", &num); | |
odd[num] ^= 1; | |
} | |
int i; | |
for (i = 1; i <= MAX_NUMBER; ++i) { | |
if (odd[i]) { | |
printf("%d\n", i); | |
exit(0); | |
} | |
} | |
return 0; | |
} |
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