Code: Select all
#include<stdio.h>
#include<math.h>
int main()
{
long int x,n,m,k,i;
scanf("%ld",&k);
for(i=0;i<k;i++)
{
scanf("%ld",&n);
{
m=(1+8*n)-1;
x=sqrt(m)/2;
printf("%ld\n",x);
}}
return 0;
}
Moderator: Board moderators
Code: Select all
#include<stdio.h>
#include<math.h>
int main()
{
long int x,n,m,k,i;
scanf("%ld",&k);
for(i=0;i<k;i++)
{
scanf("%ld",&n);
{
m=(1+8*n)-1;
x=sqrt(m)/2;
printf("%ld\n",x);
}}
return 0;
}
Code: Select all
1
100000000000000
Code: Select all
14142135
Code: Select all
#include<stdio.h>
int main()
{
long long int n, i, t, sum;
scanf("%lld",&t);
while(t--)
{
scanf("%lld",&n);
sum=0; i=1;
while(sum<n)
{
sum=sum+i;
i++;
}
if(n<sum)
{printf("%lld\n",i-2);}
else if(n==sum)
printf("%lld\n",i-1);
}
return 0;
}
I dont get your words .brianfry713 wrote:You can solve each test case in constant time.