lightoj 1294 - Positive Negative Sign 【基础计数】

最新推荐文章于 2016-08-24 09:12:37 发布

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本文介绍了一种算法挑战，即根据给定的整数n和m，将1至n的整数序列按照特定的正负号模式进行标记，并计算最终的总和。通过分析规律，提出了一种高效的解决方案。

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

12 3

4 1

Case 1: 18

Case 2: 2

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 long
using 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;
}