LightOJ 1294 - Positive Negative Sign【数学】

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
m integers 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

解题思路

有公式可以得出：将n除以二可得出有多少段，而m正好是每段的值，

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