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UVALive 6912 Prime Switch 暴力枚举+贪心

2016-09-05 10:45 477 查看

题目链接：

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4924

Prime Switch

Time limit: 1.000 seconds

问题描述

There are lamps (uniquely numbered from 1 to N) and K switches. Each switch has one prime number

written on it and it is connected to all lamps whose number is a multiple of that prime number. Pressing

a switch will toggle the condition of all lamps which are connected to the pressed switch; if the lamp

is off then it will be on, and vice versa. You can press only one switch at one time; in other words,

no two switches can be pressed together at the same time. If you want to press multiple switches, you

should do it one by one, i.e. allowing the affected lamps of the previous switch toggle their condition

first before pressing another switch.

Initially all the lamps are off. Your task is to determine the maximum number of lamps which can

be turned on by pressing one or more switches.

For example, let there be 10 lamps (1 . . . 10) and 2 switches which numbers are 2 and 5 as shown

in the following figure.

In this example:

• Pressing switch 2 will turn on 5 lamps: 2, 4, 6, 8, and 10.

• Pressing switch 5 will turn on 2 lamps: 5 and 10.

• Pressing switch 2 and 5 will turn on 5 lamps: 2, 4, 5, 6, and 8. Note that lamp number 10 will

be turned off as it is toggled twice, by switch 2 and switch 5 (off → on → off).

Among all possible switches combinations, the maximum number of lamps which can be turned on

in this example is 5.

输入

The first line of input contains an integer T (T ≤ 100) denoting the number of cases. Each case begins

with two integers in a line: N and K (1 ≤ K ≤ N ≤ 1, 000), denoting the number of lamps and

switches respectively. The next line contains K distinct prime numbers, each separated by a single

space, representing the switches number. You are guaranteed that the largest number among those

switches is no larger than N.

输出

For each case, output ‘Case #X: Y ’, where X is the case number starts from 1 and Y is the maximum

number of lamps which can be turned on for that particular case.

Explanation for 2nd sample case:

You should press switch 2 and 7, such that 11 lamps will be turned on: 2, 4, 6, 7, 8, 10, 12, 16, 18,

20, and 21. There exist some other combinations which can turn on 11 lamps, but none can turn more

than 11 lamps on.

Explanation for 3rd sample case:

There is only one switch, and pressing it will turn 20 lamps on.

Explanation for 4th sample case:

Pressing all switches will turn 42 lamps on, and it is the maximum possible in this case.

样例

sample input

4

10 2

2 5

21 4

2 3 5 7

100 1

5

100 3

3 19 7

sample output

Case #1: 5

Case #2: 11

Case #3: 20

Case #4: 42

题意

给你n盏灯，你有k个质数开关，每个质数开关可以控制是它的倍数的灯，每盏灯被打开奇数次才会亮，问如何控制开关使得亮的灯泡最多

题解

对于<31的质数开关，直接暴力枚举所有状态，对于>=31的开关（任意两个>=31的质数开关不可能共同控制同一盏灯），就贪心一下，如果把它开起来能激活更多的灯，就开，否则就关。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long  LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1111;
int n,m;

int light[maxn];
void solve(int &ret,int x){
for(int i=x;i<=n;i+=x){
light[i]^=1;
if(light[i]) ret++;
else ret--;
}
}

int main() {
int tc,kase=0;
scf("%d",&tc);
while(tc--){
scf("%d%d",&n,&m);
VI a1,a2;
rep(i,0,m){
int x; scf("%d",&x);
if(x<31) a1.pb(x);
else a2.pb(x);
}
int ans=0;
rep(i,0,(1<<a1.sz())){
clr(light,0);
int cnt=0;
rep(j,0,a1.sz()){
if((1<<j)&i){
solve(cnt,a1[j]);
}
}
rep(j,0,a2.sz()){
int tmp=0;
solve(tmp,a2[j]);
cnt+=max(0,tmp);
}
ans=max(ans,cnt);
}
prf("Case #%d: %d\n",++kase,ans);
}
return 0;
}

//end-----------------------------------------------------------------------

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