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QuestionHint and solution stay closed until you open them
Read a non-negative integer n. Print the sum 1 + 2 + … + n.
If n is 0, the sum is 0. You may use a loop or the formula n * (n + 1) / 2.
Input. One integer n.
Output. One integer — the sum.
Constraints
- 0 ≤ n ≤ 10 000
Examples
Input
5
Output
15
Input
1
Output
1
Hint
- A while loop: total = 0; i = 1; while (i <= n) { total += i; i++; }
- The closed form n * (n + 1) / 2 also works. Watch integer overflow only if n is huge — it is not, here.
Show correct code
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#include <stdio.h>
int main(void) {
int n;
if (scanf("%d", &n) == 1) {
long total = 0;
for (int i = 1; i <= n; i++) {
total += i;
}
printf("%ld\n", total);
}
return 0;
}
main.cC17 · gcc · Ctrl + Enter runs
ResultIdle
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