NYOJ436 sum of all integer numbers
2014-02-21 00:17
417 查看
原题链接
这题坑了不少人,要考虑n为负的情况。好在不管它是正是负总归是等差数列吧,这样规律就明显了,直接用(首项 + 末项)* 项数 / 2;
由于n的范围在10000之内,所以不需要考虑溢出的情况。
关键是项数 = 首尾差的绝对值 + 1;
这题坑了不少人,要考虑n为负的情况。好在不管它是正是负总归是等差数列吧,这样规律就明显了,直接用(首项 + 末项)* 项数 / 2;
由于n的范围在10000之内,所以不需要考虑溢出的情况。
关键是项数 = 首尾差的绝对值 + 1;
#include <cstdio> #include <cstdlib> int main(){ int n; while(scanf("%d", &n) == 1) printf("%d\n", (abs(n - 1) + 1) * (1 + n) / 2); return 0; }
相关文章推荐
- NYOJ - 436 - sum of all integer numbers(注意a<0)
- NYOJ436 sum of all integer numbers
- NYOJ 436 sum of all integer numbers
- NYOJ 436 sum of all integer numbers(坑人,递归)
- 436 sum of all integer numbers
- NYOJ-sum of all integer numbers
- NYOJ 46 sum of all integer numbers
- 南阳acm 436 sum of all integer numbers
- sum of all integer numbers
- sum of all integer numbers
- sum of all integer numbers
- 南阳理工:sum of all integer numbers
- sum of all integer numbers(坑题,水题)
- Given an integer, return all sequences of numbers that sum to it
- ACM-sum of all integer numbers
- sum of all integer numbers
- sum of all integer numbers
- [LeetCode] Print All Combinations of a Number as a Sum of Candidate Numbers
- [LeetCode] Print All Combinations of a Number as a Sum of Candidate Numbers
- Problem 21 of Evaluate the sum of all the amicable numbers under 10000.