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How we could get three numbers without ordering and then check if they form a Pythagorean triple or not?
So, pythagorean(3, 4, 5) or pythagorean(5, 3, 4) will print/return true, while pythagorean(4, 3, 6) will print/return false.
You can use this algotrithm :
#include<stdio.h>
int main(){
long long int a, b, c ;
scanf("%llu %llu %llu", &a, &b, &c);
if (a*a==b*b+c*c || b*b==a*a+c*c || c*c==a*a+b*b)
{
printf("YES");
}
else
printf("NO");
return 0;
}
If you use the equation, a^2 + b^2 = c^2, c should be the largest number and the order of a and b should not matter. Just find the largest number, set that equal to c, and then set the other two values to a and b and check to see if the equality is true.
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#include<stdio.h>
int main(){
int a,b[101],i,j,sum=0;
scanf("%d",&a);
for(i=0;i<a;i++){
for(j=0;j<3;j++){
scanf("%d",&b[j]);
sum=sum+b[j];
}
}
if(sum==0) printf("YES\n");
else printf("NO\n");
return 0;
}
outputs are showed perfectly but online judge didn't accept.
online judge message"wrong answer on test 81".
link of problem"http://codeforces.com/problemset/problem/69/A"
The problem is asking to check if the sum of all vectors is zero.
Sum of vectors is element-wise addition of vectors, not simple summation of all elements.
Your program will fail in, for example, this case:
1
1 0 -1
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I am trying to research various integer overflow scenarios in C and I was wondering does the C language provide any defenses against numeric errors and are there any additional classes or libraries in the C language that can help with that? Also, can anyone give me an example of code that results in an integer overflow in C?
No, there are no defenses.
This overflows:
#include <limits.h>
#include <stdio.h>
const int a = INT_MAX - 2;
const int b = INT_MAX - 2;
printf("%d + %d = %d\n", a, b, a + b);
When I tested it it printed -6, but anything could happen I guess.
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is it possible to use same integers in multiple scanf's? For example, I input int i and j, then give them a value in scanf, and print their sum. Then use another scanf to assign different values to the same integers, and now add THEIR sum..
Do you mean this?
int a, b;
scanf("%d %d", &a, &b);
printf("%d\n", a + b);
scanf("%d %d", &a, &b);
printf("%d\n", a + b);
Of course it would work. The variable's value simply changes. Its the same if you wrote
int a;
a = 4;
. . .
a = 8
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if the range of d is 1<= d <= 10^101 and n is 1<= n <= 200. since the range of double is 2.3E-308 to 1.7E+308. when i take input 11111111111111111111 as d then the value d become 11111111111111111000.000 when i show the value to terminal. that means that it couldn't take the input correctly then how will it give correct value for 10^101. i need know the nth root of d. d will be always in form of p = k^n. that's why i added pow function to know the nth root. but the problem is that the range of p is huge. what i am trying is to solve this problem Power of Cryptography !
int main(){
double d,n;
scanf("%lf%lf", &n, &d))
{
printf("%lf\n", pow(d, 1/n));
}
return 0;
}
A double precision number is not capable of holding all the values between 2.3E-308 to 1.7E+308, it is capable of holding a value between these numbers to a precision of about 15 decimal places.
That means some numbers (such as your example) require more precision than the 8 bytes of data can store.
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I just wonder how to find mean of two numbers without using division.
do not use these conditions :
int mean = (a + b) >> 1;
four fundamental arithmetic operations
I think this may be helpful -->
int a,b,i,j;
if (a>b)
{
int temp = a;
a = b;
b = temp;
}
for(i=a,j=b;i<j;i++,j--)
continue;
if(i==j)printf("%d\n", i);
else printf("%lf\n", (double)(i)-0.5);
Add them then multiply by 0.5 , no division involved.
If they're both integers, you can use a right shift:
int median = (a + b) >> 1;