poj 2668 -- Defending Castle解题报告

poj 2668 -- Defending Castle解题报告

原文：http://blog.csdn.net/zsc09_leaf/archive/2011/03/30/6289856.aspx
Description
The kingdom Eintagra is in great danger! Overwhelming enemy has surrounded the emperor's castle and once they enter, a massacre is just what is going to be. 

Now what all people in the kingdom Eintagra can rely on, is a huge catapult that can throw heavy rocks to the crowd of enemy. The catapult is so huge that it is too hard to adjust the direction it targets. So the damage it can do to the enemy is decreasing for every throw because the enemy in the targeted area are going to move away. If on the first attack it can make a certain damage, then on the second it can do only half, and 1/3 the damage on the third attack, and this holds, by estimation, that the K-th attack does 1/K damage of the first attack can do. People are optimistic so if the damage is not an integer, they round it up to the nearest bigger integer. 

Given the damage of the first attack of the catapult and the life of the catapult, people need to know how much total damage the catapult can do to the enemy. 


Input
There are multiple test cases. Each contains two positive integers D and N in a single line, where D is the damage of the first attack of the catapult, and N is its life measured by the number of attacks it can make. D and N are both positive integers and not more than 2000000000. 

Input ends with two zeros and this line should not be processed.

Output
Output a single line with an integer reporting the total damage that the catapult can do to the enemy.

Sample Input
1 1
2 3
0 0

Sample Output
1
4

Source
Beijing 2005 Preliminary 
Author:nealzane

文章很废话

说的就是已知d，n 求解d/1+d/2+d/3+....+d/n并且每个数都取上整

一开始读到这题没什么想法，想了好一会儿

举例:

d=7,n=7

则数列为

7 4 3 2 2 2 1

1 2 3 4 5 6 7

可以发现一个规律

d/i 向上取整的值表示的是 d/k=i的起始位置

例如： 7/1=7,  则7/k=1, k从7开始

又比如 7/2=4, 则 7/k=2 , k从4开始

因此我们可以获得一个 d/k=i 的一个区间

#include <iostream>
using namespace std;
int main()
{
    __int64 sum=0,b=0,s=0,e=0,num,d,n,i;
    while(scanf("%I64d%I64d",&d,&n),d+n!=0)
    {
        sum=0;
        i=1;
        e=n;                //终点初始化为n
        while(1)
        {
            s=(d-1)/i+1;    //求出值为 i的起始位置s
            sum+=s;

            if(s==i)        //避免s=e造成重复加
                s++;
            if(s<=n)
            {
                e=n<e?n:e;   //判断有效区间
                sum+=(e-s+1)*i; //加上相应的值
            }
            e=s-1;          //更新下个区间的末端
            if(e<=i||i>=n)    //跳出条件
                break;
            i++;
        }
        printf("%I64d\n",sum);
    }
}