HDU 2103 水

Family planning

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8757    Accepted Submission(s): 2267

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

 

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
long long powe(int n);
int main()
{
int T=0;
scanf("%d",&T);
while(T--)
{
int n=0,m=0;
long long sum=0;
int a[50]={0};
scanf("%d%d",&m,&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
if(n<m)
{
for(int i=1;i<=n;i++)
{
if(a[i])
{
sum=n-i;
break;
}
}
}
else
{
for(int i=1;i<=m;i++)
{
if(a[i])
{
sum=n-i;
break;
}
}
if(sum==0)
{
sum=n-m;
}
}
long long ans=0;
ans=(powe(sum)-1)*10000;
printf("%lld RMB\n",ans);
}
return 0;
}
long long powe(int n)
{
    long long sum=1;
    for(int i=1;i<=n;++i)
        sum*=2;
    return sum;
}