B1012. Digital classification

Title Description

　　Given some of the positive integer column, please be classified according to the requirements of digital, and digital outputs the following five categories:

• A1 = divisible numbers can be 5 and all even
• A2 = 5 can be more than 1 after the digital addition is performed in the order given staggered sum, i.e. calculate n1 - n4 ... - n2 + n3
• A3 = 2 is the number of the digital I except 5
• A4 = 5 is the average addition to more than 3 after the numbers to the nearest decimal place
• A5 = 4 is the maximum number than the number 5 in addition to

Input Format

　　Each input comprises a test, each test is given to a positive integer not more than 1000 N, and then gives the N does not exceed 1000 to be classified positive integer. Separated by spaces between numbers

Output Format

　　For a given positive integer N, is calculated according to A1 ~ A5 subject of the request and sequentially outputting in a row. Between numbers separated by spaces, but the end of the line may not have extra spaces

　　If a certain type wherein the number does not exist, the corresponding output position "N"

Sample input 1

13 1 2 3 4 5 6 7 8 9 10 20 16 18

Sample Output 1

30 11 2 9.7 9

Sample input 2

8 1 2 4 5 6 7 9 15

Sample Output 2

N 11 2 N 9

int main(int argc, char *argv[]) {
    int res[5] = {0, 0, 0, 0, 0};
    int count[5] = {0, 0, 0, 0, 0};
    int n, temp;
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%d", &temp);
        if(temp % 5 == 0){
            if(temp % 2 == 0){
                years [ 0 ] + Temp;
                count[0]++;            
            }
        }else if(temp % 5 == 1){
            if(count[1] % 2 == 0){
                res[1] += temp;
            }else{
                res[1] -= temp;
            }
        }else if(temp % 5 == 2){
            res[2]++;
        }else if(temp % 5 == 3){
            res[3] += temp;
            count[3]++;
        }else if(temp % 5 == 4){
            if(count[4] < temp){
                count[4] = temp;
            }
            count[4]++;
        }
    } 
    if(count[0] == 0){
        printf("N ");
    }else{
        printf("%d ", res[0]);
    }
    if(count[1] == 0){
        printf("N ");
    }else{
        printf("%d ", res[1]);
    }
    if(count[2] == 0){
        printf("N ");
    }else{
        printf("%d ", res[2]);
    }
    if(count[3] == 0){
        printf("N ");
    }else{
        printf("%.1f ", (double)res[3]/count[3]);
    }
    if(count[4] == 0){
        printf("N");
    }else{
        printf("%d", res[4]);
    }
    return 0;
}

The key problem solution

• Filled pit format, one decimal place, .1f