Problem:
Score : 100 points

Problem Statement

Two students of AtCoder Kindergarten are fighting over candy packs.

There are three candy packs, each of which contains a, b, and c candies, respectively.

Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.

Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Constraints

1 ≦ a, b, c ≦ 100

Input

The input is given from Standard Input in the following format:

a b c

Output

If it is possible to distribute the packs so that each student gets the same number of candies, print Yes. Otherwise, print No.

Sample Input 1

10 30 20

Sample Output 1

Yes

Give the pack with 30 candies to one student, and give the two packs with 10 and 20 candies to the other. Then, each gets 30 candies.

Sample Input 2

30 30 100

Sample Output 2

No

In this case, the student who gets the pack with 100 candies always has more candies than the other.

Note that every pack must be given to one of them.

Sample Input 3

56 25 31

Sample Output 3

Yes

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
	int array[100][100];
	int W,H,N;
	int x,y,a;
	int i,j,k;
	int sum;
	
	scanf("%d %d %d",&W,&H,&N);
	for(k=0;k<N;k++){
		scanf("%d %d %d",&x,&y,&a);
		switch(a){
			case 1:
				if(x!=0){
					for(j=0;j<x;j++){
						for(i=0;i<H;i++)array[j][i]=1;
					}
				}
				break;
			case 2:
				if(x!=W){
					for(j=x;j<W;j++){
						for(i=0;i<H;i++)array[j][i]=1;
					}
				}
				break;
			case 3:
				if(y!=0){
					for(i=0;i<y;i++){
						for(j=0;j<W;j++)array[j][i]=1;
					}
				}
				break;
			case 4:
				if(y!=H){
					for(i=y;i<H;i++){
						for(j=0;j<W;j++)array[j][i]=1;
					}
				}
				break;
		}
	}
	for(i=0;i<H;i++){
		for(j=0;j<W;j++){
			sum+=array[j][i];
		}
	}
	printf("%d\n",W*H-sum);
	return 0;
}