【lightoj1294】数学
来源:互联网 发布:西安鼓楼网络售票 编辑:程序博客网 时间:2024/05/18 06:27
Description
Given two integers: n and m and n is divisible by 2m, you have to write down the first n natural numbers in the following form. At first take first mintegers and make their sign negative, then take next m integers and make their sign positive, the next m integers should have negative signs and continue this procedure until all the n integers have been assigned a sign. For example, let n be 12 and m be 3. Then we have
-1 -2 -3 +4 +5 +6 -7 -8 -9 +10 +11 +12
If n = 4 and m = 1, then we have
-1 +2 -3 +4
Now your task is to find the summation of the numbers considering their signs.
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case starts with a line containing two integers: n and m (2 ≤ n ≤ 109, 1 ≤ m). And you can assume that n is divisible by 2*m.
Output
For each case, print the case number and the summation.
Sample Input
2
12 3
4 1
Sample Output
Case 1: 18
Case 2: 2
规律题,我开始还想着打表,后来发现规律了轻松ac
#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define LL long longusing namespace std;//void init() {//for(int i=1; i<n; i++) {//a[i]=a[i-1]+i;//}//}int main() {int T,p=0;scanf("%d",&T);while(T--) {LL n,m;scanf("%lld%lld",&n,&m);LL cnt=0;cnt=n/(2*m)*m;printf("Case %d: %lld\n",++p,cnt*m);}return 0;}
- 【lightoj1294】数学
- lightoj1294 - Positive Negative Sign
- LightOj1294 Positive Negative Sign
- Lightoj1294——Positive Negative Sign(神坑)
- 数学
- 数学
- 数学
- 数学
- 数学
- 数学
- 数学
- 数学
- 数学
- 数学
- 数学
- 数学
- 数学
- 数学
- 两学一做
- 字符串池和堆
- TOJ 1438.Perfect Pth Powers
- 823oracle
- ReactNative设置 Image容器的 圆角背景
- 【lightoj1294】数学
- python多线程
- 几种硬盘IO性能测试工具
- AndroidCallPhone
- 数据结构实验之排序四:寻找大富翁
- Rocketmq 消息的解压缩
- 数据库相关
- MySQL索引原理及慢查询优化
- 最小的k个数