20 | 9月 | 2015 | 超客的水墨烟雨

poj1014

maksyuki 发表于 oj 分类，标签: DP-背包

20 9月 2015

Dividing

Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value. Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.

Input

Each line in the input file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , n6 , where ni is the number of marbles of value i. So, the example from above would be described by the input-line "1 0 1 2 0 0". The maximum total number of marbles will be 20000.
The last line of the input file will be "0 0 0 0 0 0"; do not process this line.

Output

For each collection, output "Collection #k:", where k is the number of the test case, and then either "Can be divided." or "Can't be divided.".
Output a blank line after each test case.

Sample Input

1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0

Sample Output

Collection #1:

Can't be divided.

Collection #2:

Can be divided.

Source

Mid-Central European Regional Contest 1999

题目类型：多重背包

算法分析：使用dp[i]表示恰好组成体积i时所具有的最大价值，递推方程为dp[i] = max (dp[i], dp[i-w[i]] + w[i])，初始化为dp[0] = 0，其他都为-1，最后判断dp[v] == v是否成立即可

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 2e6 + 66;

//c[i]表示第i个物品的花费 w[i]表示第i个物品的价值
//m[i]表示第i个物品的数量 dp[i]表示前i物品组成的最大价值
int c[maxn], w[maxn], m[maxn], dp[maxn];
int V, N;//分别表示背包的总体积和可选的物品的种数

void ZeroOnePack (int c, int w)
{
    for (int i = V; i >= c; i--)
        dp[i] = max (dp[i], dp[i-c] + w);
}

void CompletePack (int c, int w)
{
    for (int i = c; i <= V; i++)
        dp[i] = max (dp[i], dp[i-c] + w);
}

void MultiPack (int c, int w, int m)
{
    if (c * m >= V)
    {
        CompletePack (c, w);
        return ;
    }

    int k = 1;
    while (k < m)
    {
        ZeroOnePack (k * c, k * w);
        m -= k;
        k *= 2;
    }
    ZeroOnePack (m * c, m * w);
}

void Solve ()//可依照题目变化，这里表示背包不需要装满，如果只有dp[0]赋值为0，则表示恰好装满时的最大价值
{
    memset (dp, -1, sizeof (dp));
    dp[0] = 0;
    for (int i = 1; i <= N; i++)
        MultiPack (c[i], w[i], m[i]);
}


int main()
{
//    CFF;
    N = 6;
    int flag = 1;
    while (scanf ("%d%d%d%d%d%d", &m[1], &m[2], &m[3], &m[4], &m[5], &m[6]) != EOF)
    {
        V = 0;
        for (int i = 1; i <= 6; i++)
        {
            V += i * m[i];
            c[i] = w[i] = i;
        }

        if (!V)
            break;

        printf ("Collection #%d:\n", flag++);

        if (V % 2) puts ("Can't be divided.\n");

        else
        {
            V /= 2;
            Solve ();

            if (dp[V] == V)
                puts ("Can be divided.\n");

            else
                puts ("Can't be divided.\n");
        }
    }
    return 0;
}

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#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 2e6 + 66;

//c[i]表示第i个物品的花费 w[i]表示第i个物品的价值

//m[i]表示第i个物品的数量 dp[i]表示前i物品组成的最大价值

int c[maxn], w[maxn], m[maxn], dp[maxn];

int V, N;//分别表示背包的总体积和可选的物品的种数

void ZeroOnePack (int c, int w)

{

for (int i = V; i >= c; i--)

dp[i] = max (dp[i], dp[i-c] + w);

}

void CompletePack (int c, int w)

{

for (int i = c; i <= V; i++)

dp[i] = max (dp[i], dp[i-c] + w);

}

void MultiPack (int c, int w, int m)

{

if (c * m >= V)

{

CompletePack (c, w);

return ;

}

int k = 1;

while (k < m)

{

ZeroOnePack (k * c, k * w);

m -= k;

k *= 2;

}

ZeroOnePack (m * c, m * w);

}

void Solve ()//可依照题目变化，这里表示背包不需要装满，如果只有dp[0]赋值为0，则表示恰好装满时的最大价值

{

memset (dp, -1, sizeof (dp));

dp[0] = 0;

for (int i = 1; i <= N; i++)

MultiPack (c[i], w[i], m[i]);

}

int main()

{

// CFF;

N = 6;

int flag = 1;

while (scanf ("%d%d%d%d%d%d", &m[1], &m[2], &m[3], &m[4], &m[5], &m[6]) != EOF)

{

V = 0;

for (int i = 1; i <= 6; i++)

{

V += i * m[i];

c[i] = w[i] = i;

}

if (!V)

break;

printf ("Collection #%d:\n", flag++);

if (V % 2) puts ("Can't be divided.\n");

else

{

V /= 2;

Solve ();

if (dp[V] == V)

puts ("Can be divided.\n");

else

puts ("Can't be divided.\n");

}

return 0;

}

2015ACM-ICPC亚洲区北京站网赛(1/10)

maksyuki 发表于比赛分类，

20 9月 2015

H 描述

This is the logo of PKUACM 2016. More specifically, the logo is generated as follows:

Put four points A₀(0,0), B₀(0,1), C₀(1,1), D₀(1,0) on a cartesian coordinate system.
Link A₀B₀, B₀C₀, C₀D₀, D₀A₀separately, forming square A₀B₀C₀D₀.
Assume we have already generated square A_iB_iC_iD_i, then square A_i+1B_i+1C_i+1D_i+1is generated by linking the midpoints of A_iB_i, B_iC_i, C_iD_iand D_iA_isuccessively.
Repeat step three 1000 times.

Now the designer decides to add a vertical line x=k to this logo( 0<= k < 0.5, and for k, there will be at most 8 digits after the decimal point). He wants to know the number of common points between the new line and the original logo.

输入

In the first line there’s an integer T( T < 10,000), indicating the number of test cases.

Then T lines follow, each describing a test case. Each line contains an float number k, meaning that you should calculate the number of common points between line x = k and the logo.

输出

For each test case, print a line containing one integer indicating the answer. If there are infinity common points, print -1.

样例输入

0.375

0.001

0.478

样例输出

-1

题目类型：二分查找

算法分析：由题目可知内部的每个小正方形都可以看作是相对较大的正方形二分边得来的，此时可以对于每个输入的坐标在[0,0. 5)二分坐标值

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 3e5 + 66;

int PP (double val)
{
    double a = 0, b = 0.5, m;
    int res = 0;
    while (b - a > EPS)
    {
        m = (a + b) / 2;
        if (fabs (m - val) < EPS)
            return -1;

        else if (val + EPS < m)
        {
            res += 4;
            break;
        }

        else
        {
            res += 4;
            a = m;
        }
    }
    return res;
}

int main()
{
//    CPPFF;
    int t;
    cin >> t;
    while (t--)
    {
        double val;
        cin >> val;
        if (fabs (val - 0) < EPS)
            cout << "-1" << endl;

        else
            cout << PP (val) << endl;
    }
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 3e5 + 66;

int PP (double val)

{

double a = 0, b = 0.5, m;

int res = 0;

while (b - a > EPS)

{

m = (a + b) / 2;

if (fabs (m - val) < EPS)

return -1;

else if (val + EPS < m)

{

res += 4;

break;

}

else

{

res += 4;

a = m;

}

return res;

}

int main()

{

// CPPFF;

int t;

cin >> t;

while (t--)

{

double val;

cin >> val;

if (fabs (val - 0) < EPS)

cout << "-1" << endl;

else

cout << PP (val) << endl;

}

return 0;

}

BestCoder Round #56(Div.2) (0/4) (Div.1) (0/4)

maksyuki 发表于比赛分类，标签: 比赛-bestcoder

20 9月 2015

超客的水墨烟雨

poj1014

2015ACM-ICPC亚洲区北京站网赛(1/10)

H 描述

输入

输出

BestCoder Round #56(Div.2) (0/4) (Div.1) (0/4)

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