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2013
12-29

hdu 2061 Treasure the new start, freshmen![解题报告]C++

Treasure the new start, freshmen!

问题描述 :

background:
A new semester comes , and the HDU also meets its 50th birthday. No matter what’s your major, the only thing I want to tell you is:"Treasure the college life and seize the time." Most people thought that the college life should be colorful, less presure.But in actual, the college life is also busy and rough. If you want to master the knowledge learned from the book, a great deal of leisure time should be spend on individual study and practise, especially on the latter one. I think the every one of you should take the learning attitude just as you have in senior school.
"No pain, No Gain", HDU also has scholarship, who can win it? That’s mainly rely on the GPA(grade-point average) of the student had got. Now, I gonna tell you the rule, and your task is to program to caculate the GPA.
If there are K(K > 0) courses, the i-th course has the credit Ci, your score Si, then the result GPA is
GPA = (C1 * S1 + C2 * S2 +……+Ci * Si……) / (C1 + C2 + ……+ Ci……) (1 <= i <= K, Ci != 0)
If there is a 0 <= Si < 60, The GPA is always not existed.

输入:

The first number N indicate that there are N test cases(N <= 50). In each case, there is a number K (the total courses number), then K lines followed, each line would obey the format: Course-Name (Length <= 30) , Credits(<= 10), Score(<= 100).
Notice: There is no blank in the Course Name. All the Inputs are legal

输出:

The first number N indicate that there are N test cases(N <= 50). In each case, there is a number K (the total courses number), then K lines followed, each line would obey the format: Course-Name (Length <= 30) , Credits(<= 10), Score(<= 100).
Notice: There is no blank in the Course Name. All the Inputs are legal

样例输入:

2
3
Algorithm 3 97
DataStruct 3 90
softwareProject 4 85
2
Database 4 59
English 4 81

样例输出:

90.10

Sorry!

#include<iostream>
using namespace std;
char ch[35];

int main()
{
	int n, m, k = 0;
	scanf("%d", &n);
	while(n--)
	{
		scanf("%d", &m);
		int i;
		double s = 0, count = 0;
		bool p = true;
		for(i = 0; i < m; i++)
		{
			double a, b;
			scanf("%s%lf%lf", ch, &a, &b);
			if(b < 60)
				p = false;
			s += a*b;
			count += a;
		}
		if(k)
			printf("\n");
		k = 1;
		if(p == false)
			printf("Sorry!\n");
		else
			printf("%.2lf\n", s / count);
	}
	return 0;
}

解题转自:http://blog.csdn.net/asure__cpp/article/details/8592832


  1. I like your publish. It is great to see you verbalize from the coronary heart and clarity on this essential subject matter can be easily noticed.

  2. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。

  3. 您没有考虑 树的根节点是负数的情况, 若树的根节点是个很大的负数,那么就要考虑过不过另外一边子树了