Run ID | Author | Problem | Lang | Verdict | Time | Memory | Code Length | Submit Time |
---|---|---|---|---|---|---|---|---|
3868 | 毕潇天 | [在线测评解答教程] 求和 | C | Accepted | 0 MS | 188 KB | 126 | 2023-01-11 11:34:04 |
#include<stdio.h> int main() { int n; while (scanf("%d", &n) != EOF) { printf("%d\n", n * (n + 1) / 2 ); } return 0; }