## Tags » Devskill Solution

#### DCP-421: Divisible Pairs [solved]

```/******************************************
Mobarak Hosen Shakil
ICE, Islamic University
ID: mhiceiuk(all), 29698(LOJ)
E-mail: mhiceiuk @ (Gmail/Yahoo/FB)
Blog: https://iuconvergent.wordpress.com
*******************************************/

#include<bits/stdc++.h>
using namespace std;

int main()
{
int tn, cn=0;
scanf("%d", &tn);
while(tn--)
{
int n;
scanf("%d", &n);
long long arr[6], a;
memset(arr, 0, sizeof(arr));
for(int i=0; i<n; i++)
{
scanf("%d", &a);
arr++;
}

long long int ans=0,b,c,d;
a=b=c=d=0;
if(arr[0]>1) a=(arr[0]*(arr[0]-1))/2;
if(arr[3]>1) b=(arr[3]*(arr[3]-1))/2;
c=(arr[5]*arr[1]);
d=(arr[4]*arr[2]);
ans=a+b+c+d;
printf("%lld\n",ans);
}
}
```
Devskill Solution

#### DCP-303: Fibonacci Divisor

```/******************************************
Mobarak Hosen Shakil
ICE, Islamic University
ID: mhiceiuk(all), 29698(LOJ)
E-mail: mhiceiuk @ (Gmail/Yahoo/FB)
Blog: http://iuconvergent.wordpress.com
*******************************************/

#include<cstdio>
using namespace std;

int main()
{
//IOS;
int tn, n, k;
scanf("%d", &tn);
while(tn--)
{
scanf("%d%d", &n, &k);
k/=n;
printf("%d\n", k);
}
return 0;
}
```
Devskill Solution

#### DCP-251: Journey By Circle

```/******************************************
Mobarak Hosen Shakil
ICE, Islamic University
ID: mhiceiuk(all), 29698(LOJ)
E-mail: mhiceiuk @ (Gmail/Yahoo/FB)
Blog: https://iuconvergent.wordpress.com
*******************************************/

#include<cstdio>
using namespace std;

template<typename T>inline T Gcd(T a, T b){return b?Gcd(b, a%b):a;}
template<typename T>inline T Lcm(T a, T b){return (a*b)/Gcd(a, b);}

int main()
{

int tn;
long long int r1, r2;
scanf("%d", &tn);
while(tn--)
{
scanf("%lld%lld", &r1, &r2);
printf("%lld\n", Lcm(r1, r2)*2);
}
return 0;
}
```
Math

#### DCP-185: Maximum Difference

```/******************************************
Mobarak Hosen Shakil
ICE, Islamic University
ID: mhiceiuk(all), 29698(LOJ)
E-mail: mhiceiuk @ (Gmail/Yahoo/FB)
Blog: https://iuconvergent.wordpress.com
*******************************************/

#include<cstdio>
#include<cmath>
using namespace std;
#define Max(a, b) a<b?b:a

int main()
{

int tn, tc=1, a1, a2, b1, b2;
scanf("%d", &tn);
while(tn--){
scanf("%d%d%d%d", &a1, &b1, &a2, &b2);

int mx=0;
mx=Max(abs(a1-a2), abs(b1-b2));
mx=Max(mx, abs((a1+b2)-(a2+b1)));
printf("Case %d: %d\n", tc++, mx);

}
return 0;
}
```
Devskill Solution

#### DCP-165: Easy Geometry [solved]

```/******************************************
Mobarak Hosen Shakil
ICE, Islamic University
ID: mhiceiuk(all), 29698(LOJ)
E-mail: mhiceiuk @ (Gmail/Yahoo/FB)
Blog: https://iuconvergent.wordpress.com
*******************************************/

#include<iostream>
#include<cstdio>
using namespace std;

int main()
{
int tn, tc=1;
scanf("%d", &tn);

while(tn--)
{
int nrow, ncol;
scanf("%d%d", &nrow, &ncol);
printf("Case %d: %d\n", tc++, nrow*2+ncol*3);
}
return 0;
}
```
Math

#### DCP-138: The Last Fibonacci [solved]

```/******************************************
Mobarak Hosen Shakil
ICE, Islamic University
ID: mhiceiuk(all), 29698(LOJ)
E-mail: mhiceiuk @ (Gmail/Yahoo/FB)
Blog: https://iuconvergent.wordpress.com
*******************************************/

#include<bits/stdc++.h>
using namespace std;
#define MX 100000
#define MX2 10000
int arr, arr2;
int main()
{
int n, tn, tc=1;
arr[0]=0;
arr[1]=1;
for(int i=2; i<=60; i++)
arr[i]=(arr+arr)%10;
scanf("%d", &tn);
while(tn--)
{
scanf("%d", &n);
printf("Case %d: ", tc++);
if(n<61) printf("%d", arr[n]);
else printf("%d", arr);
printf(" is the last digit.\n");
}
return 0;
}
```
Devskill Solution

#### DCP-23: Another Bigmod Problem [solved]

```/******************************************
Mobarak Hosen Shakil
ICE, Islamic University
ID: mhiceiuk(all), 29698(LOJ)
E-mail: mhiceiuk @ (Gmail/Yahoo/FB)
Blog: https://iuconvergent.wordpress.com
*******************************************/

#include <iostream>
using namespace std;
#define ll long long int
template <typename T>
T mmul(T a, T b, T m) {
a %= m;
T result = 0;
while (b) {
if (b % 2) result = (result + a) % m;
a = (a + a) % m;
b /= 2;
}
return result;
}

template <typename T>
T mpow(T a, T b, T m) {
a %= m;
T result = 1;
while (b) {
if (b % 2) result = mmul(result, a, m);
a = mmul(a, a, m);
b /= 2;
}
return result;
}
int  main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int tn, tc=0;
cin>>tn;
while(tn--)
{
ll a, p, m, res;
cin>>a>>p>>m;
res=mpow(a, p, m);
cout<<"Case "<<++tc<<": "<<res<<endl;
}
return 0;
}
```
Devskill Solution