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: Runtime Error)



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main() {
    char S[100001];
    int  O[100000];
    int n,i=0,count,jmin,j,index,flag=0,nbR=0,nbL=0;
   
    
    scanf("%s",S);
    n=strlen(S);
    for (i=0;i<n;i++){
        O[i]=0;
    }
    for (i=0;i<n;i++){
           if ((S[i]=='R')){
              nbR++;
            }
           if ((S[i]=='L')){
              nbL++;
            }
           if ((S[i]=='R')&&(S[i+1]=='L')){
             index=i; 
           }
           else if (((S[i]=='L')&&(S[i+1]=='R'))||(i==n-1)){
               if ((nbR%2==1) && (nbL%2==0)){
                 count=nbR+nbL;
                 O[index]=count/2;
                 O[index+1]=count-O[index];
               }
               else {
                   count=nbR+nbL;
                   O[index+1]=count/2;
                   O[index]=count-O[index+1];
               }
               nbR=0;
               nbL=0;
               
           }
           
    } 
    for (i=0;i<n;i++){
        printf("%d\n",O[i]);
    }
    return (EXIT_SUCCESS);
}