Linux - 操作文件与目录(manipulating files and directories)
浏览数：131 /
时间：2015年06月20日
Cards
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1214 Accepted Submission(s): 265
Problem Description
Given some cards each assigned a number, you‘re required to select EXACTLY K cards among them.
While you select a card, I will check the number assigned to it and see if it satisfies some of the following conditions:
1. the number is a prime number;
2. the amount of its divisors is a prime number;
3. the sum of its divisors is a prime number;
4. the product of all its divisors is a perfect square number. A perfect square number is such a kind of number that it can be written as a square of an integer.
The score you get from this card is equal to the amount of conditions that its number satisfies. The total score you get from the selection of K cards is equal to the sum of scores of each card you select.
After you have selected K cards, I will check if there‘s any condition that has never been satisfied by any card you select. If there is, I will add some extra scores to you for each unsatisfied condition. To make the game more interesting, this score may be negative.
After this, you will get your final score. Your task is to figure out the score of each card and find some way to maximize your final score.
Note that 1 is not a prime number. In this problem, we consider a number to be a divisor of itself. For example, considering the number 16, it is not a prime. All its divisors are respectively 1, 2, 4, 8 and 16, and thus, it has 5 divisors with a sum of 31 and a product of 1024. Therefore, it satisfies the condition 2, 3 and 4, which deserves 3 points.
While you select a card, I will check the number assigned to it and see if it satisfies some of the following conditions:
1. the number is a prime number;
2. the amount of its divisors is a prime number;
3. the sum of its divisors is a prime number;
4. the product of all its divisors is a perfect square number. A perfect square number is such a kind of number that it can be written as a square of an integer.
The score you get from this card is equal to the amount of conditions that its number satisfies. The total score you get from the selection of K cards is equal to the sum of scores of each card you select.
After you have selected K cards, I will check if there‘s any condition that has never been satisfied by any card you select. If there is, I will add some extra scores to you for each unsatisfied condition. To make the game more interesting, this score may be negative.
After this, you will get your final score. Your task is to figure out the score of each card and find some way to maximize your final score.
Note that 1 is not a prime number. In this problem, we consider a number to be a divisor of itself. For example, considering the number 16, it is not a prime. All its divisors are respectively 1, 2, 4, 8 and 16, and thus, it has 5 divisors with a sum of 31 and a product of 1024. Therefore, it satisfies the condition 2, 3 and 4, which deserves 3 points.
Input
The first line of the input contains the number of test cases T.
Each test case begins with two integers N and K, indicating there are N kinds of cards, and you‘re required to select K cards among them.
The next N lines describes all the cards. Each of the N lines consists of two integers A and B, which denote that the number written on this kind of card is A, and you can select at most B cards of this kind.
The last line contains 4 integers, where the ith integer indicates the extra score that will be added to the result if the ith condition is not satisfied. The ABSOLUTE value of these four integers will not exceed 40000.
You may assume 0<N≤103,0<K≤104,1≤A≤106,1≤B≤104,T≤40 and the total N of all cases is no more than 20000. In each case there are always enough cards that you‘re able to select exact K cards among them.
Each test case begins with two integers N and K, indicating there are N kinds of cards, and you‘re required to select K cards among them.
The next N lines describes all the cards. Each of the N lines consists of two integers A and B, which denote that the number written on this kind of card is A, and you can select at most B cards of this kind.
The last line contains 4 integers, where the ith integer indicates the extra score that will be added to the result if the ith condition is not satisfied. The ABSOLUTE value of these four integers will not exceed 40000.
You may assume 0<N≤103,0<K≤104,1≤A≤106,1≤B≤104,T≤40 and the total N of all cases is no more than 20000. In each case there are always enough cards that you‘re able to select exact K cards among them.
Output
Output two lines for each test case.
The first line consists of N integers separated by blanks, where the ith integer is the score of the ith card.
The second line contains a single integer, the maximum final scores you can get.
The first line consists of N integers separated by blanks, where the ith integer is the score of the ith card.
The second line contains a single integer, the maximum final scores you can get.
Sample Input
1 5 3 1 1 2 1 3 1 4 1 5 1 1 2 3 4
Sample Output
1 3 2 2 2 11
Source
思路来源于：cxlove博客
题意：
每个数，有4个评估
1、是素数 2、约数个数是素数 3、约数的和是素数 4、约数的乘积是完全平方数
从n个中选出k个，使得分数最高，如果选的k个中没有一个满足某个条件，则有另外的分值。
思路：
sqrt枚举n的约数，打一个大点的素数表能判断1、2、3；至于4的话考虑到n有约数i时同时也有n/i，而i*n/i=n，故可以边乘边处理，注意n自身是平方数的情况i和n/i只取一个，预处理完了之后就是判断怎样得出最佳答案了。
数只有可能有2^4种状态，而每种状态的得分又是确定的，可以统一处理，暴力枚举每种数是否出现，然后根据每一种出现的状态，先将出现的数全部取一个（使取得数满足这种状态），然后从得分高的往得分低的取（贪心），如果能恰好取k个，加上附加值，更新答案就够了。
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 10005
#define MAXN 200005
#define OO (1<<31)-1
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int n,m,ans,cnt,tot,flag;
int a,b,state,sc;
int s[5],score[16];
bool vis[4000005];
struct Node
{
int s,num;
}xx[20];
void sieve(int nn) // 筛选素数
{
int i,j;
memset(vis,0,sizeof(vis));
vis[1]=1;
for(i=2; i<=nn; i++)
{
if(nn/i<i) break ;
if(!vis[i])
{
for(j=i*i; j<=nn; j+=i)
{
vis[j]=1;
}
}
}
}
bool cmp(Node x,Node y)
{
return x.num>y.num;
}
void init()
{
int i,j,t,num;
sieve(4000000);
for(i=0;i<16;i++) // 得到状态和每种状态的得分
{
xx[i].s=i;
t=i; num=0;
while(t>0)
{
if(t&1) num++;
t>>=1;
}
xx[i].num=num;
}
sort(xx,xx+16,cmp); // 按照得分排序
}
bool issquare(ll pro) // 判断是否是平方数
{
ll d=sqrt(pro*1.0);
if((d-1)*(d-1)==pro||d*d==pro||(d+1)*(d+1)==pro) return true ;
return false ;
}
void solve()
{
int i,j,t,flg,num,sum;
ll pro=1;
num=sum=0;
for(i=1; i*i<=a; i++) // sqrt(a)枚举约数
{
if(a%i==0)
{
pro*=i;
if(i*i!=a)
{
num+=2; sum+=i+a/i; pro*=a/i;
}
else
{
num++; sum+=i;
}
if(pro==ll(a)*ll(a)) pro=1; // 注意这里的强制转换！
}
}
if(!vis[a]) flg=1;
else flg=0;
state=sc=0;
if(flg) sc++,state|=1;
if(!vis[num]) sc++,state|=(1<<1);
if(!vis[sum]) sc++,state|=(1<<2);
if(issquare(pro)) sc++,state|=(1<<3);
score[state]+=b;
}
int main()
{
int i,j,t;
init();
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
memset(score,0,sizeof(score));
vector<int>v;
for(i=1; i<=n; i++)
{
scanf("%d%d",&a,&b);
solve();
v.push_back(sc);
}
for(i=0; i<4; i++)
{
scanf("%d",&s[i]);
}
tot=1<<16;
ans=-INF;
int res,k,ns;
for(i=1;i<tot;i++) // 每一种数出现的状态
{
res=ns=0; k=m;
for(j=0;j<16;j++) // 每一种状态的数 先全部取一个
{
if(!(i&(1<<j))) continue ;
ns|=xx[j].s;
res+=xx[j].num;
k--;
if(score[xx[j].s]<1) k=-INF; // 该状态没有值
}
if(k<0) continue ; // 取的超出k个
for(j=0;j<16;j++) // 然后贪心 从分高的往低的取
{
if(!(i&(1<<j))) continue ;
if(k>score[xx[j].s]-1)
{
k-=score[xx[j].s]-1;
res+=(score[xx[j].s]-1)*xx[j].num;
}
else
{
res+=k*xx[j].num;
k=0;
break ;
}
}
if(k==0) // 如果能够取k个
{
for(j=0;j<4;j++)
{
if(!(ns&(1<<j))) res+=s[j];
}
ans=max(ans,res);
}
}
for(i=0;i<n;i++)
{
if(i!=0) printf(" ");
printf("%d",v[i]);
}
printf("\n%d\n",ans);
}
return 0;
}
Linux - 操作文件与目录(manipulating files and directories),古老的榕树,5-wow.com
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