【LightOJ 1294 Positive Negative Sign】

Positive Negative Sign

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

#include<cstdio>int main(){    int T,pl=1;    long long ans,N,M,kl,a;    scanf("%d",&T);    while(T--)    {        scanf("%lld%lld",&N,&M);        if(N%M==0)        {            if(N%2==0)            ans=M*M*(N/M/2);            else            {                ans=M*M*(N/M/2);                ans-=(2*N-M+1)*M/2;            }        }        else        {           kl=N%M;           N-=kl;           if(N%2==0)           {               ans=M*M*(N/M/2);               ans-=(2*N+kl)*kl/2;           }           else           {               ans=M*M*(N/M/2);               ans+=(2*N+kl)*kl/2;               ans-=(2*N-M+1)*M/2;           }        }        printf("Case %d: %lld\n",pl++,ans);    }    return 0;}FAQ | About | Google Group | Discuss | Author All Copyright Reserved ©2010-2016 HUST ACM/ICPC TEAM