【lightoj1294】数学-蒲公英云

【lightoj1294】数学

墨蓝 2022-07-17 00:27 167阅读 0赞

H - H 使用long long

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1294 uDebug

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

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 long
using 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;
}