C题解:
n = 1 | n = 2 | n = 3 | n = 4 | n = 5 |
---|---|---|---|---|
2 | 5 | 8 | 11 | 14 |
规律:
value = 3 * n - 1
2到3n-1的求和:
(2 + (3n - 1)) * n / 2 = n * (3n + 1) / 2
因此题解如下:
int main() {
int n = 0;
while(scanf("%d", &n) != -1) {
int sum = n * (3 * n + 1) / 2;
printf("%d\n", sum);
}
return 0;
}