解题思路:对ans = {∑(1≤i≤n) i }+ 1。每次切割都与前i - 1刀有交点的情况下是最大的。
#includelong long sum; int n; int main () { while (scanf("%d", &n), n > = 0) { sum = 1; for (long long i = 1; i <= n; i++) sum += i; printf("%lld\n", sum); } return 0; }
#includelong long sum; int n; int main () { while (scanf("%d", &n), n > = 0) { sum = 1; for (long long i = 1; i <= n; i++) sum += i; printf("%lld\n", sum); } return 0; }