UVA10300（理解题意就容易切）

Ecological Premium

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18810
Description

German farmers are given a premium depending on the conditions at their farmyard. Imagine the following simplified regulation: you know the size of each farmer's farmyard in square meters and the number of animals living at it. We won't make a difference between different animals, although this is far from reality. Moreover you have information about the degree the farmer uses environment-friendly equipment and practices, expressed in a single integer greater than zero. The amount of money a farmer receives can be calculated from these parameters as follows. First you need the space a single animal occupies at an average. This value (in square meters) is then multiplied by the parameter that stands for the farmer's environment-friendliness, resulting in the premium a farmer is paid per animal he owns. To compute the final premium of a farmer just multiply this premium per animal with the number of animals the farmer owns.
Input
The first line of input contains a single positive integer n (<20), the number of test cases. Each test case starts with a line containing a single integer f (0<f<20), the number of farmers in the test case. This line is followed by one line per farmer containing three positive integers each: the size of the farmyard in square meters, the number of animals he owns and the integer value that expresses the farmer’s environment-friendliness. Input is terminated by end of file. No integer in the input is greater than 100000 or less than 0.

Output
For each test case output one line containing a single integer that holds the summed burden for Germany's budget, which will always be a whole number. Do not output any blank lines.

Sample Input
3
5
1 1 1
2 2 2
3 3 3
2 3 4
8 9 2
3
9 1 8
6 12 1
8 1 1
3
10 30 40
9 8 5
100 1000 70
Sample Output
38
86

7445

解题思路：

这道题的题意刚开始没看懂，仔细读发现题意是这样的。给你n个farmer，然后每个farmer给出占地（a）、动物数(b)、环境友好参数(c)。问你最后所有farmer的总收益。

它的计算规则是这样的 : a / b算出每个动物的占地，然后 * c ，再 * b ，这样的出每个farmer的收益，最后做下和就好了。但是推导到最后其实就是每个farmer的a * c 值，取下和就行。

完整代码：

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;typedef double DB;typedef unsigned uint;typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const DB EPS = 1e-9;const DB OO = 1e20;const DB PI = acos(-1.0); //M_PI;struct node{    int a , b  , c;}s[1111];int main(){    #ifdef DoubleQ    freopen("in.txt","r",stdin);    #endif    int T;    cin >> T;    while(T--)    {        int n;        cin >> n;        LL sum = 0;        for(int i = 0 ; i < n ; i ++)        {            cin >> s[i].a >> s[i].b >> s[i].c;            sum += s[i].a * s[i].c;        }        cout << sum << endl;    }}