The Mussels

Time Limit: 1000ms
Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 2439
64-bit integer IO format: %I64d      Java class name: Main

 
"To be or not to be, that is the question." Now FJ( Frank&John) faces a serious problem.

FJ is breeding mussels these days. The mussels all want to be put into a culturist with a grade no little than their own grades. FJ accept their requirements.

FJ provides culturists of different grades, all having a certain capacity. FJ first put mussels into culturists with the same grade until they are full. Then he may put some mussels into some potential culturists that still have capacity.

Now, FJ wants to know how many mussels can be put into the culturists.

 

Input

For each data set:
The first line contains two integers, n and m(0<n<=100000,0<m<=1000000), indicating the number of culturists and the number of mussels. The ith culturist has a grade i, and grade 1 is considered the highest.

The second line contains n integers indicating the culturists' capacity in order.

The third line contains m integers all in the range 1~n, indicating the mussels' grade in order.

Proceed to the end of file.

 

Output

A single integer, which is the number of mussels that can be put into the culturists.

 

Sample Input

4 4
100 4 4 4
1 2 3 4
4 3
1 1 1 1
4 2 2

Sample Output

4
3

Source

 
解题:贪心+模拟,直接按照题目意思去做就是了。。。
 
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
int cul[],tmp,n,m;
void myscanf(int &x){
char ch;
while((ch = getchar()) < '' || ch > '');
x = ;
x = x* + ch - '';
while((ch = getchar()) >= '' && ch <= '') x = x* + ch - '';
}
int main() {
while(~scanf("%d %d",&n,&m)){
for(int i = ; i <= n; i++)
myscanf(cul[i]);
int ans = ;
for(int i = ; i <= m; i++){
myscanf(tmp);
if(cul[tmp]){
--cul[tmp];
++ans;
}else{
for(int j = tmp; j; --j){
if(cul[j]){
--cul[j];
++ans;
break;
}
}
}
}
printf("%d\n",ans);
}
return ;
}

HDU 2439 The Mussels的更多相关文章

  1. HDOJ 2111. Saving HDU 贪心 结构体排序

    Saving HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total ...

  2. 【HDU 3037】Saving Beans Lucas定理模板

    http://acm.hdu.edu.cn/showproblem.php?pid=3037 Lucas定理模板. 现在才写,noip滚粗前兆QAQ #include<cstdio> #i ...

  3. hdu 4859 海岸线 Bestcoder Round 1

    http://acm.hdu.edu.cn/showproblem.php?pid=4859 题目大意: 在一个矩形周围都是海,这个矩形中有陆地,深海和浅海.浅海是可以填成陆地的. 求最多有多少条方格 ...

  4. HDU 4569 Special equations(取模)

    Special equations Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u S ...

  5. HDU 4006The kth great number(K大数 +小顶堆)

    The kth great number Time Limit:1000MS     Memory Limit:65768KB     64bit IO Format:%I64d & %I64 ...

  6. HDU 1796How many integers can you find(容斥原理)

    How many integers can you find Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d ...

  7. hdu 4481 Time travel(高斯求期望)(转)

    (转)http://blog.csdn.net/u013081425/article/details/39240021 http://acm.hdu.edu.cn/showproblem.php?pi ...

  8. HDU 3791二叉搜索树解题(解题报告)

    1.题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=3791 2.参考解题 http://blog.csdn.net/u013447865/articl ...

  9. hdu 4329

    problem:http://acm.hdu.edu.cn/showproblem.php?pid=4329 题意:模拟  a.     p(r)=   R'/i   rel(r)=(1||0)  R ...

随机推荐

  1. HDU 2586 How far away ?(LCA模板 近期公共祖先啊)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 Problem Description There are n houses in the vi ...

  2. 关于SharePoint讨论板的一些知识(2)--视图中的栏目

    关于SharePoint讨论板的一些知识(2)--视图中的栏目         新建讨论后,默认显示四个栏目:主题.创建者.答复和上次更新时间.         从功能区的当前视图能够看出这是默认的主 ...

  3. Java高级程序猿技术积累

    watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveDczNDQwMDE0Ng==/font/5a6L5L2T/fontsize/400/fill/I0JBQk ...

  4. ZOJ 1654 Place the Robots(最大匹配)

    Robert is a famous engineer. One day he was given a task by his boss. The background of the task was ...

  5. PHP统计所有字符在字符串中出现的次数

    <?php //统计字符串中出现的字符,出现次数 echo '<pre>'; $str = 'aaabbccqqwweedfghhjffffffffggggggggg';//字符串示 ...

  6. Cocos2d-x《雷电大战》(3)-子弹无限发射

    林炳文Evankaka原创作品.转载请注明出处http://blog.csdn.net/evankaka 本文要实现雷电游戏中,游戏一開始,英雄飞机就无限发射子弹的功能. 这里的思想是单独给子弹弄一个 ...

  7. Linux - 常用网络命令详解netstat,scp

    ifconfig 查看生效的ip信息. [root@local ~]# ifconfig eno16777736: flags=4163<UP,BROADCAST,RUNNING,MULTICA ...

  8. [POJ 2282] The Counting Problem

    [题目链接] http://poj.org/problem?id=2282 [算法] 数位DP [代码] #include <algorithm> #include <bitset& ...

  9. 南海区行政审批管理系统接口规范v0.3(规划) 3.业务办理API 3.1.businessAuditById【业务办理】

    {"c_accept":"Q2015112400002","c_operators":"gz99","v_op ...

  10. git 本地项目推送至远程仓库

    1 在本地文件夹下创建一个 Git 仓库(如test目录下) git init 2 此时test文件夹即是你的maste主分支,你可以在改文件夹下写自己的项目 3 将test文件夹下的内容提交至暂存区 ...