【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.

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

5410 ACM 杭电 01+完全背包
题目:http://acm.hdu.edu.cn/showproblem.php?pid=5410 虽然是英文题目:但还是很好理解的.明显的背包问题思路:如果你能想到把题目拆分成小问题,就会简单许多 ...
最详细最简单：最大公因数求法、辗转相除法、更相减损法，入门ACM，杭电水题，算法递归，初级算法题一看就懂
文章目录前言一.名称定义 1.最大公约数 2.辗转相除法 3.更相减损法二.ACM杭电入门题 1.解题思路三.解题参考代码(C语言,C++) 0.最优算法(C++) 1.辗转相除求解(C语言) ...
杭电OJ分类题目（2）
原题出处:HDOJ Problem Index by Type,http://acm.hdu.edu.cn/typeclass.php 杭电OJ分类题目(2) HDU Water~~~ HDU 100 ...
杭电ACM（HDUOJ）试题分类
杭电ACM试题分类第一篇 1001 这个就不用说了吧 1002 简单的大数 1003 DP经典问题,最 ...
杭电acm第2304题答案c语言,【转】杭电ACM试题分类
注:网上搜的第一篇 1001 这个就不用说了吧 1002 简单的大数 1003 DP经典问题,最大连续子段和 1004 简单题 1005 找规律(循环点) 1006 感觉有点BT的题,我到现在还没过 ...
杭电ACM题目类型整理
版权声明:(╯3╰) 转载请注明: http://blog.csdn.net/bat67 杭电acm题目分类版本1 1002 简单的大数 1003 DP经典问题,最大连续子段和 1004 简单题 10 ...
【杭电ACM】1097 A hard puzzle
[杭电ACM]1097 A hard puzzle http://acm.hdu.edu.cn/showproblem.php?pid=1097 先用int手写了算法结果竟然wrong answer ...
杭电ACM刷题（1）：1002，A + B Problem II
最近忙于考试复习,没有多少可供自己安排的时间,所以我利用复习之余的空闲时间去刷刷杭电acm的题目,也当对自己编程能力的锻炼吧. Problem Description I have a very si ...
杭电acm2151题java的解法,杭电ACM题型分类
杭电ACM题型分类基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029.1032.1037.1040.1048.1 ...

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

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

最新文章

热门文章