lightOJ 1294 - Positive Negative Sign 【规律题】

1294 - Positive Negative Sign

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
2 12 3 4 1	Case 1: 18 Case 2: 2

PROBLEM SETTER: JANE ALAM JAN

思路：

刚开始还以为是一道模拟题，然后直接超时，然后仔细一看，原来是一道规律题！
思路1： 由于题上说是n能被2*m整除，所以说，n/m后得到的是偶数，那么也就是后面的m组数，减去前面的m组数得到的是m*m，然后看有多少个m*m,也就是将n/m得到的结果除以2就是m*m的个数，然后相乘就是想要的结果！
思路2：直接将n/2代表的是后面那个数加上前面那个负数所得的值（也就是m）的个数，然后用n/2*m就是结果了！
记住用long long型，否则会出错！

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
typedef long long ll;
using namespace std;
ll n,m;
int main()
{
int T;
scanf("%d",&T);
int N=T;
while(T--)
{
scanf("%lld%lld",&n,&m);
printf("Case %d: ",N-T);
printf("%lld\n",n/2*m);
}
return 0;
}