lightoj 1294 - Positive Negative Sign 【基础计数】

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

 

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PROBLEM SETTER: JANE ALAM JAN

题意：给你1-n共n个数，给你一个数m(n % (2*m) == 0)，让你每次截取m个数，按 [- + - +...]的顺序给它们添加符号，最后得到一个表达式，问表达式的结果是多少。

思路：首先n为10^9，不能模拟。

考虑 [- +] 临近2*m个数做出的贡献，求得值为m*m。同样若考虑[+ -] 临近2*m个数做出的贡献，求得值为-m*m。

这样，只需求出n分成多少个m，设temp = n / m，则相应的[- +] 或 [+ -]个数为temp / 2。

若temp为偶数，结果为所有[- +]做出的贡献。反之结果为所有[+ -]做出的贡献 + 前m个数之和。 

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <vector>#define INF 0x3f3f3f3f#define eps 1e-8#define MAXN 500000+10#define MAXM 50000000#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while(a--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long longusing namespace std;int main(){    int t, kcase = 1;    Ri(t);    W(t)    {        LL n, m;        Rl(n); Rl(m);        LL temp = n / m;        LL sum;        if(temp & 1)        {            temp--;            sum = -m * (m + 1)/2;            sum += -m * m * temp / 2;        }        else            sum = m * m * temp / 2;        printf("Case %d: %lld\n", kcase++, sum);    }    return 0;}