uva 10079 - Pizza Cutting
2013-10-26 12:45
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题目链接:uva 10079 - Pizza Cutting
题目大意:有一个披萨,切n刀,问最多可以切几块。
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。
题目大意:有一个披萨,切n刀,问最多可以切几块。
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。
#include <stdio.h> long 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; }
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