Chef and Dolls

Chef is fan of pairs and he likes all things that come in pairs. He even has a doll collection in which all dolls have paired. One day while going through his collection he found that there are odd number of dolls. Someone had stolen a doll!!!

Help chef find which type of doll is missing/

Input

The first line contains the number of test cases.
Second line of the input contains the number of elements in the array.
The next n lines are the types of each doll that is left.

Output

Find the type of doll that doesn't have a pair

Constraints

1<=T<=10
1<=N<=100000 (10^5)
1<=ti<=100000

#include <stdio.h>

int main()

{   int i,j,c,z,y,x[100000],k,a[100000],b,n,m,t;

    scanf("%d",&t);

    for(i=0;i<t;i++)

    {   scanf("%d",&n);

        for(j=0;j<n;j++)

            scanf("%d",&a[j]);

        y=0;

        for(j=0;j<n;j++)

        {   b=a[j];

            c=0;

            for(m=0;m<n;m++)

                        if(a[m]==b)

                            c++;

            if(c%2!=0)

            {   for(z=0;z<=y;z++)

                    if(x[z]==b)

                        break;

                    else

                        printf("%d\n",b);

                x[y]=b;

            }

        }

    }

    return 0;

}

Id and Ship

Write a program that takes in a letter class ID of a ship and display the equivalent string class description of the given ID. Use the table below.

Class ID	Ship Class
B or b	Battle Ship
C or c	Cruiser
D or d	Destroyer
F or f	Frigate

Input

The first line contains an integer T, total number of test cases. Then follow T lines, each line contains a character.

Output

Display the Ship Class depending on ID.

Constraints

1 ≤ T ≤ 1000

#include <stdio.h>

int main()

{   int n,i;

    char ch;

    scanf("%d",&n);

    for(i=0;i<2*n;i+=1)

    {   scanf("%c",&ch);

        switch(ch)

        {   case 'b': printf("BattleShip\n");

                    break;

            case 'B': printf("BattleShip\n");

                    break;

            case 'c': printf("Cruiser\n");

                    break;

            case 'C': printf("Cruiser\n");

                    break;

            case 'd': printf("Destroyer\n");

                    break;

            case 'D': printf("Destroyer\n");

                    break;

            case 'f': printf("Frigate\n");

                    break;

            case 'F': printf("Frigate\n");

                    break;

        }  

    }  

    return 0;

}

First and Last Digit

If Give an integer N, Write a program to obtain the sum of the first and last digit of this number.

Input

The first line contains an integer T, total number of test cases. Then follow T lines, each line contains an integer N.

Output

Display the sum of first and last digit of N.

Constraints

1 ≤ T ≤ 1000

1 ≤ N ≤ 1000000

#include <stdio.h>

int main()

{   int t,i,n,x;

    scanf("%d",&t);

    for(i=0;i<t;i++)

    {   scanf("%d",&n);

        if(n<10)

            printf("%d",n);

        else

        {   x=n%10;

            while(n>=10)

                n=n/10;

            x+=n;

            printf("%d\n",x);

        }

    }

    return 0;

}

Valid Triangles

Write a program to check whether a triangle is valid or not, when the three angles of the triangle 
are the inputs. A triangle is valid if the sum of all the three angles is equal to 180 degress.

Input

The first line contains an integer T, total number of testcases. Then follow T lines, each line contains three angles A, B and C of triangle separated by space.

Output

Display 'YES' or 'NO' if the triangle is Valid or not respectively.

Constraints

1 ≤ T ≤ 1000

40 ≤ A,B,C ≤ 180

#include <stdio.h>

int main()

{   int t,i,n,x,y,z;

    scanf("%d",&t);

    for(i=0;i<t;i++)

    {   scanf("%d",&x);

        scanf("%d",&y);

        scanf("%d",&z);

        if(((x+y+z)==180)&&(x>0)&&(y>0)&&(z>0))

            printf("YES\n");

        else

            printf("NO\n");

    }

    return 0;

}