【ACM】杭电1178:Heritage from father 小心溢出！

我觉得做这个题 最大的意义就是通过NNNN次的WA让我反复而深刻地体会到了 计算过程中也会溢出 这种错误。。。。

另外要把一个数用科学计数法表示，只需要两步：

1、对这个数取以10为底的对数，其结果取整就是科学计数法中10需要的指数。

2、用这个数除以 【10 ^ (步骤一的结果)】就是科学计数法的有效数字。

代码：

#include <stdio.h>
#include <math.h>


/* Sn = n(n+1)(n+2)/6; */

int main(int argc, char *argv[])
{
	int N;
	while(scanf("%d",&N) != EOF && N != 0)
	{
	/*	double sum = N * (N + 1) / 6.0 * (N + 2);  第一句开始我是这样写的，结果老是WA .注意，N * (N + 1) 这个结果会溢出 */
		double sum = N * ((N + 1) / 6.0) * (N + 2); /* 因此要先计算 (N + 1) / 6.0 ，这样数就小了，不会溢出了 */
		int bit = (int)log10(sum);
		double head = sum / pow(10,bit);
		
		printf("%.2lfE%d\n",head,bit);
	}
	return 0;
}

Problem Description

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.

 

Input

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.

 

Output

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)

 

Sample Input

1
3
0

 

Sample Output

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.