首页 > ACM题库 > HDU-杭电 > HDU 1178 Heritage from father[解题报告] C++
2013
12-03

HDU 1178 Heritage from father[解题报告] C++

Heritage from father

问题描述 :

Famous Harry Potter,who seemd to be a normal and poor boy,is actually a wizard.Everything changed when he had his birthday of ten years old.A huge man called ‘Hagrid’ found Harry and lead him to a new world full of magic power.
If you’ve read this story,you probably know that Harry’s parents had left him a lot of gold coins.Hagrid lead Harry to Gringotts(the bank hold up by Goblins). And they stepped into the room which stored the fortune from his father.Harry was astonishing ,coz there were piles of gold coins.
The way of packing these coins by Goblins was really special.Only one coin was on the top,and three coins consisted an triangle were on the next lower layer.The third layer has six coins which were also consisted an triangle,and so on.On the ith layer there was an triangle have i coins each edge(totally i*(i+1)/2).The whole heap seemed just like a pyramid.Goblin still knew the total num of the layers,so it’s up you to help Harry to figure out the sum of all the coins.

输入:

The input will consist of some cases,each case takes a line with only one integer N(0<N<2^31).It ends with a single 0.

输出:

对于每个输入的N,输出一行,采用科学记数法来计算金币的总数(保留三位有效数字)

样例输入:

1
3
0

样例输出:

1.00E0
1.00E1

Hint
Hint
when N=1 ,There is 1 gold coins. when N=3 ,There is 1+3+6=10 gold coins.

额,直接用公式吧

还有,科学计数法没办法直接按要求输出,得直接求出指数

#include <stdio.h>
int main ()
{
int n,c;
double sum;
while (scanf("%d",&n)!=EOF)
{
   if (n==0)break;
   sum=1.0/6*n*(n+2)*(n+1);
   c=0;
   while (sum>=10)
   {
    c++;
    sum/=10;
   }
   printf ("%.02lfE%d\n",sum,c);
}
return 0;
}

  1. 这道题目虽然简单,但是小编做的很到位,应该会给很多人启发吧!对于面试当中不给开辟额外空间的问题不是绝对的,实际上至少是允许少数变量存在的。之前遇到相似的问题也是恍然大悟,今天看到小编这篇文章相见恨晚。