HDU2103Family planning

Family planning

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9139    Accepted Submission(s): 2362

Problem Description

As far as we known,there are so many people in this world,expecially in china.But many people like LJ always insist on that more people more power.And he often says he will burn as much babies as he could.Unfortunatly,the president XiaoHu already found LJ's extreme mind,so he have to publish a policy to control the population from keep on growing.According the fact that there are much more men than women,and some parents are rich and well educated,so the president XiaoHu made a family planning policy:
According to every parents conditions to establish a number M which means that parents can born M children at most.But once borned a boy them can't born other babies any more.If anyone break the policy will punished for 10000RMB for the first time ,and twice for the next time.For example,if LJ only allowed to born 3 babies at most,but his first baby is a boy ,but he keep on borning another 3 babies, so he will be punished for 70000RMB(10000+20000+40000) totaly.

 

Input

The first line of the input contains an integer T(1 <= T <= 100) which means the number of test cases.In every case first input two integers M(0<=M<=30) and N(0<=N<=30),N represent the number of babies a couple borned,then in the follow line are N binary numbers,0 represent girl,and 1 represent boy.

 

Output

Foreach test case you should output the total money a couple have to pay for their babies.

 

Sample Input

  
  

   
   2
2 5
0 0 1 1 1

2 2
0 0
  
  

 

题目比较简单，但还是WA了好多次，最后 发现是输出结果处没有加上“ RMB”，也是醉了；

题目大意；

大概就是说国家通过计划生育来控制人口和男女比例；超生的是要被处以罚款的；

国家的计划生育规定每个人家里最多可以生 M 个孩子，超出的孩子就要按 第一个罚款 10000，第二个罚款 20000， 第三个罚款 40000 的等比数列依次递增下去；每个人家里有 N 个孩子，每超出一个就按上面的罚款标准罚款，如果这个人家里生的孩子在最多的允许范围内但是这些孩子中有男的，则从这个男的孩子开始，后面的都算是超生，都是要罚钱的；

给出AC代码；

#include<iostream>
using namespace std;

int main()
{
	int t, m, n;
	cin >> t;
	while (t--)
	{
		cin >> m >> n;
		long long sum = 0, num = 10000;
		int flag = 0;
		for (int i = 1; i <= n; i++)
		{
			int  sex;
			cin >> sex;
			
			if (i>m||flag)
			{
				sum += num;
				num *= 2;
			}
			if (sex)flag = 1;
			//cout << "flag=" << flag << endl;
			//cout << "i=" << i << endl;
		}
		cout << sum << " RMB" << endl;
	}
	return 0;
}