Family planning

在草稿箱呆了好几年了，才放出来

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11294 Accepted Submission(s): 2971

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

Sample Output

70000 RMB 0 RMB

题目要求：本题的大概意思就是求不按规则生娃要交多少罚款？规则有两条：1.不能超生。 2.即使没有超生，但生完男娃后就不能再生啦。（注意第一个男娃是免费的，我由于忽略了这个问题，结果第一次提交wa掉）

c语言代码：（在草稿箱呆了好几年了，才放出来）

#include<stdio.h>
int main()
{
int t,m,n,i,a;
scanf("%d",&t);
while(t--)
{
long long int sum=0,ss=10000,l=0,ll=0;
scanf("%d%d",&m,&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a);
if(ll==0&&i<=m&&a==1)
{ll=1;continue;
}
if(i<=m&&ll>0)  l++;               //计算有多少个不和规矩的娃
if(i>m)    l++;                           //这是超生的娃
}
for(i=1;i<=l;i++)                             //计算罚款
{
sum+=ss;
ss*=2;
}
printf("%lld RMB\n",sum);
}
return 0;
}

杭电2103---Family planning相关推荐

hdoj杭电问题分类
杭电上的题虽然多,但是一直苦于找不到问题分类,网页都是英文的,所以平时做题也没怎么看,今天仔细一看,问题分类竟然就在主页....做了那么久的题居然没发现,表示已经狗带..不要笑,不知道有没有像我一样傻 ...
杭电acm第2304题答案c语言,【转】杭电ACM试题分类
注:网上搜的第一篇 1001 这个就不用说了吧 1002 简单的大数 1003 DP经典问题,最大连续子段和 1004 简单题 1005 找规律(循环点) 1006 感觉有点BT的题,我到现在还没过 ...
杭电acm2151题java的解法,杭电ACM题型分类
杭电ACM题型分类基础题:1000.1001.1004.1005.1008.1012.1013.1014.1017.1019.1021.1028.1029.1032.1037.1040.1048.1 ...
杭电OJ分类题目（3）
原题出处:HDOJ Problem Index by Type,http://acm.hdu.edu.cn/typeclass.php 杭电OJ分类题目(3) HDU Computational Ge ...
杭电ACM（HDUOJ）试题分类
杭电ACM试题分类第一篇 1001 这个就不用说了吧 1002 简单的大数 1003 DP经典问题,最 ...
杭电46道DP牛人总结
原文地址:杭电46道DP牛人总结作者:飞泉鸣玉杭电46道DP牛人总结 HDU 动态规划(46道题目)倾情奉献~ [只提供思路与状态转移方程] 收藏 Robberies http://acm.hdu. ...
杭电ACM题目类型整理
版权声明:(╯3╰) 转载请注明: http://blog.csdn.net/bat67 杭电acm题目分类版本1 1002 简单的大数 1003 DP经典问题,最大连续子段和 1004 简单题 10 ...
【杭电ACM】1097 A hard puzzle
[杭电ACM]1097 A hard puzzle http://acm.hdu.edu.cn/showproblem.php?pid=1097 先用int手写了算法结果竟然wrong answer ...
【ACM】杭电OJ 1106 函数atoi
函数atoi是把字符串转化成整数的函数,头文件为 #include "stdlib.h" e.g. 运行环境:Dev-C++ 5.11 杭电1106 调用了sort函数,运行的时间 ...
【ACM】杭电OJ 2037
题目链接:杭电OJ 2037 先把b[i]进行排序,然后,b[i]与a[i+1]进行比较. #include <iostream> #include <cstdio> #inc ...

杭电2103---Family planning

Family planning

杭电2103---Family planning相关推荐

最新文章

热门文章