ACM程序设计选修课——1051: Glamor Sequence(YY+求和公式)

1051: Glamor Sequence

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 16   Solved: 5
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Description

you have a sequence +1, +2, +3, ... +m, -(m + 1), -(m + 2), ..., -(2m), (2m + 1), (2m + 2), ..., (3m), .....
and you need calculator sum of the first n items.

Input

For each test case, there is a line contains two integers n and m, (1 <= m <= n <= 1000000000).

Output

For each test case, print sum of the first n items.

Sample Input

8 2
10 3

Sample Output

-8
5

HINT

For the first sample, the sequence is 1, 2, -3, -4, 5, 6, -7, -8, so the answer is -8.

For the second sample, the sequence is 1, 2, 3, -4, -5, -6, 7, 8, 9, -10, so the answer is 5.

想了很久还是不会。问了同学大概思路,确是用等差数列求和公式,但是先前的做法效率比较低,用了for循环,一旦出现n比较大而m非常小的话就会超时,比如n=1000000000,m=1,test测试一下5秒多。

因此一种思路是这样:

可以发现从1开始前2*m项的和的绝对值是一个常数即m,e=n-n%(2*m)那么只需处理后面e+1~n的数字即可。因此可以有如下规律

设剩下的区间为e+1~n(也可能是e+m)。

1、若e+m<=n——即数轴分布情况为e+1~e+m~n。只需将之前的所有完整段的和加上Sum[e+1,e+m]即可,且每一项都是正数。

2、若e+m>n——即数轴分布情况为e+1~n~e+m。那么要计算S1[e+1,n]与S2[n+1,e+m],ans=ans+S1-S2。

#include<iostream> 
#include<string> 
#include<algorithm> 
using namespace std; 
int main(void) 
{ 
    long long n,m,e; 
    long long ans; 
    while (~scanf("%lld%lld",&n,&m)) 
    { 
        e=n-n%(2*m); 
        ans=(e)*(-m)/2;//所有完整段求和
        if(e+m<n)//分类讨论
        {
        	ans=ans+(2*e+1+m)*m/2;
        	ans=ans-(e+m+1+n)*(n-e-m)/2;
        }
        else if(e+m>n)
        {
        	ans=ans+(e+1+n)*(n-e)/2;
        }
        else
        {
        	ans=ans+(2*e+1+m)*m/2;
        }
    	printf("%lld\n",ans);//这题要用long long来储存答案
    } 
    return 0; 
}   

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