how to printf each result of looping

Question

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Print sum of N integer numbers.

Input

The first line contains integer 1 ≤ T ≤ 100, denoting the number of test cases. Each case start with a number N, and the second line contains N integers separated by one space character.

Output

T lines, in each line, print sum of input numbers.

Sample Input

C#

2
5
54 78 0 4 9
3
1 2 3

Sample Output

C#

145
6

I want make printf all answer at the end..
this what I tried :

C#

#include<stdio.h>
int main()
{
    int T, N, test, array[20], sum[20];
    do {
        printf ("How Many Test Case \n");
        scanf ("%d", &T);
        if (T<1 || T>100){
            printf ("Please Input Integer 0<T<100 \n");
        }
    }    while (T<1 || T>100);

    for (test=0; test<T ; test++){
        printf ("How many Integers to sum? \n");
        scanf ("%d", &N);
        int i=0;
        while(i<N)
        scanf("%d",&array[i++]);

        int m=0, sum=0;
        for (m=0; m<N; m++)
        {
            sum = sum + array[m];
            sum = sum[j]
        }
    for (x=0; x<T; x++)
    printf ("%d\n", sum[j]);

        }
    return 0;
    }

Posted 14-May-15 4:24am

Member 11690687

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Comments

Frankie-C 14-May-15 10:54am

And the problem is?
While you explain better to us what the problem is take note of:
- Undefined 'j' and 'x' variables.
- Missing ';' at end of line: sum = sum[j]
- sum redefined locally on line: int m=0, sum=0;

2 solutions

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chandanadhikari · Answer 1 · 2015-05-14T04:57:00

hi,
code is mostly correct.
It seems the problem is in this part :

C++

for (m=0; m<N; m++)//calling this loop1
{
    sum = sum + array[m];
    sum = sum[j]
}

for (x=0; x<T; x++)//calling this loop2
    printf ("%d\n", sum[j]);

first of all i do not see j defined anywhere.
suggestion ::

C++

#include<stdio.h>
int main()
{
    int T, N, test, array[20], sum[20];
	//declare another variable to store sum temporarily
	int tempSum;
    do
	{
        printf ("How Many Test Case \n");
        scanf ("%d", &T);
        if (T<1 || T>100){
            printf ("Please Input Integer 0<T<100 \n");
        }
    }    while (T<1 || T>100);
 
    for (test=0; test<T ; test++)
	{
        printf ("How many Integers to sum? \n");
        scanf ("%d", &N);
        int i=0;
        while(i<N)
        scanf("%d",&array[i++]);
 
        int m=0, tempSum=0;
        for (m=0; m<N; m++)
        {
            tempSum = tempSum + array[m];
        }
		sum[test] = tempSum;
	}	
		for (x=0; x<T; x++)
		printf ("%d\n", sum[x]);
 
     return 0;
    }

did not test this by executing so please test yourself.
hope it works !!

Frankie-C · Answer 2 · 2015-05-14T05:16:00

This is the correct code:

C++

 #include<stdio.h>
int main()
{
    int T, N, test, array[20], sum[100];
    do
	{
        printf ("How Many Test Case \n");
        scanf ("%d", &T);
        if (T<1 || T>100)
		{
            printf ("Please Input Integer 0<T<100 \n");
        }
    }    while (T<1 || T>100);
 
    for (test=0; test<T ; test++)
	{
        printf ("How many Integers to sum? \n");
        scanf ("%d", &N);
        int i=0;
        while(i<N)
        	scanf("%d",&array[i++]);
 
        int m=0;
		sum[test]=0;
        for (m=0; m<N; m++)
            sum[test] += array[m]; 
    }

	int x;
    for (x=0; x<T; x++)
    	printf ("%d\n", sum[x]);

    return 0;
}

how to printf each result of looping

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