Problem:
Score : 100 points

Problem Statement

We have two bottles for holding water.

Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.

Bottle 2 contains C milliliters of water.

We will transfer water from Bottle 2 to Bottle 1 as much as possible.

How much amount of water will remain in Bottle 2?

Constraints

All values in input are integers.

1 \leq B \leq A \leq 20

1 \leq C \leq 20

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.

Sample Input 1

6 4 3

Sample Output 1

1

We will transfer two milliliters of water from Bottle 2 to Bottle 1, and one milliliter of water will remain in Bottle 2.

Sample Input 2

8 3 9

Sample Output 2

4

Sample Input 3

12 3 7

Sample Output 3

0

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

#include <stdio.h>
int main(void)
{
int s,i,a,N;
scanf("%d",&N);
if(N==1){scanf("%d",&s);a=0;}
else{
int sa[N-1],H[N];
a=0;
for(i=1;i<N+1;i++){
 scanf("%d",&H[i-1]);
 if(i>1){
 sa[i-2]=H[i-2]-H[i-1];	
 if(sa[i-2]>=2){a=a+2;}
else if(sa[i-i]==1){a=a+1; }
else{a=a;}
}
}
}
if(a>=2){
	printf("No");
}
else{
	printf("Yes");
}
return 0;
}