HDU2103 Family planning

Family planning

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10142    Accepted Submission(s): 2632

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

22 50 0 1 1 12 20 0

 

Sample Output

70000 RMB0 RMB

 

Source

HDU 2007-6 Programming Contest

 

本题题意为M表示一对夫妇应该生的孩子的数量，N表示实际生孩子的数量，如果第一次生的为男孩，则以后出生的孩子都要交罚款，如果在M数量内的孩子里面没有男孩，则超出M外的孩子要交罚款。由于最后结果了能会溢出，所以要进行处理。

#include <stdio.h>int a[35];int main(){int t,n,m;scanf("%d",&t);while(t --){scanf("%d%d",&m,&n);int flag = 0;long long sum = 0;for(int i = 0; i < n; i ++)scanf("%d",&a[i]);int money = 1;for(int i = 0; i < n; i ++){if(flag){sum += money;money *= 2;}if(a[i] && i < m)flag = 1;if(!flag && i >= m){sum += money;money *= 2;}}if(sum)printf("%lld0000 RMB\n",sum);elseprintf("0 RMB\n");}return 0;}