zoukankan      html  css  js  c++  java
  • HDU 2124 Repair the Wall(贪心)

    Repair the Wall

    Problem Description
    Long time ago , Kitty lived in a small village. The air was fresh and the scenery was very beautiful. The only thing that troubled her is the typhoon.

    When the typhoon came, everything is terrible. It kept blowing and raining for a long time. And what made the situation worse was that all of Kitty's walls were made of wood.

    One day, Kitty found that there was a crack in the wall. The shape of the crack is 
    a rectangle with the size of 1×L (in inch). Luckly Kitty got N blocks and a saw(锯子) from her neighbors.
    The shape of the blocks were rectangle too, and the width of all blocks were 1 inch. So, with the help of saw, Kitty could cut down some of the blocks(of course she could use it directly without cutting) and put them in the crack, and the wall may be repaired perfectly, without any gap.

    Now, Kitty knew the size of each blocks, and wanted to use as fewer as possible of the blocks to repair the wall, could you help her ?
     
    Input
    The problem contains many test cases, please process to the end of file( EOF ).
    Each test case contains two lines.
    In the first line, there are two integers L(0<L<1000000000) and N(0<=N<600) which
    mentioned above.
    In the second line, there are N positive integers. The ith integer Ai(0<Ai<1000000000 ) means that the ith block has the size of 1×Ai (in inch).
     
    Output
    For each test case , print an integer which represents the minimal number of blocks are needed.
    If Kitty could not repair the wall, just print "impossible" instead.
     
    Sample Input
    5 3
    3 2 1
    5 2
    2 1
     
    Sample Output
    2
    impossible
     
    Answer
    简简单单的题意要用这么长一段去说...用一堆木头去补墙,木头可以锯断,如果不够输出impossible,否则输出最少使用的木头。先判断木头总长是否小于需修补的长度,如果小于,输出impossible,否则,从大到小排序,直到总长大于等于需修补的长度。
     
    #include <cstdio>
    #include <iostream>
    #include <string>
    #include <sstream>
    #include <cstring>
    #include <stack>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #include <map>
    #define PI acos(-1.0)
    #define ms(a) memset(a,0,sizeof(a))
    #define msp memset(mp,0,sizeof(mp))
    #define msv memset(vis,0,sizeof(vis))
    using namespace std;
    //#define LOCAL
    int cmp(int a,int b){return a>b;}
    int main()
    {
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
    #endif // LOCAL
        ios::sync_with_stdio(false);
        int wall,n;
        vector<int> v;
        while(cin>>wall>>n)
        {
            v.clear();
            int sum=0;
            for(int i=0;i<n;i++)
            {
                int t;
                cin>>t;
                sum+=t;
                v.push_back(t);
            }
            if(sum<wall)printf("impossible
    ");
            else
            {sort(v.begin(),v.end(),cmp);
            int ans=0,cnt=0;
            for(int i=0;i<(int)v.size();i++)
            {
                if(cnt>=wall)break;
                else cnt+=v[i],ans++;
            }
            printf("%d
    ",ans);}
        }
        return 0;
    }
    View Code
  • 相关阅读:
    JAVA中的多态
    JAVA中的策略模式strategy
    JAVA中的clone方法剖析
    JAVA虚拟机中的堆内存Heap与栈内存Stack
    JAVA垃圾回收分代处理思想
    JAVA 垃圾回收机制
    JAVA内存管理
    混迹于博客园很久了,今天终于有了自己的博客园:coding-of-love 嘿嘿
    小程序富文本wxParse转换不成功的解决办法,填坑
    elementui级联选择器 如何设置多选?
  • 原文地址:https://www.cnblogs.com/gpsx/p/5187010.html
Copyright © 2011-2022 走看看