| 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 |
题解：这个规律比较好推，直接每隔m一正一负，依次相减就是m了，因为整除2*m，就直接是m*n/2；因为没有加case，wa了一次，还是太不细心了啊；
代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #define mem(x,y) memset(x,y,sizeof(x)) 7 using namespace std; 8 typedef long long LL; 9 const int INF=0x3f3f3f3f; 10 int main(){ 11 int T,flot=0; 12 LL n,m; 13 scanf("%d",&T); 14 while(T--){ 15 scanf("%lld%lld",&n,&m); 16 printf("Case %d: %lld ",++flot,m*(n/2)); 17 } 18 return 0; 19 }