poj1365

maksyuki 发表于 oj 分类，标签: 数论-素因子分解

15 9月 2015

Prime Land

Everybody in the Prime Land is using a prime base number system. In this system, each positive integer x is represented as follows: Let {pi}i=0,1,2,... denote the increasing sequence of all prime numbers. We know that x > 1 can be represented in only one way in the form of product of powers of prime factors. This implies that there is an integer kx and uniquely determined integers e_kx, e_kx-1, ..., e₁, e₀, (e_kx > 0), that The sequence (e_kx, e_kx-1, ... ,e₁, e₀) is considered to be the representation of x in prime base number system. It is really true that all numerical calculations in prime base number system can seem to us a little bit unusual, or even hard. In fact, the children in Prime Land learn to add to subtract numbers several years. On the other hand, multiplication and division is very simple. Recently, somebody has returned from a holiday in the Computer Land where small smart things called computers have been used. It has turned out that they could be used to make addition and subtraction in prime base number system much easier. It has been decided to make an experiment and let a computer to do the operation minus one''. Help people in the Prime Land and write a corresponding program. For practical reasons we will write here the prime base representation as a sequence of such pi and ei from the prime base representation above for which ei > 0. We will keep decreasing order with regard to pi.

Input

The input consists of lines (at least one) each of which except the last contains prime base representation of just one positive integer greater than 2 and less or equal 32767. All numbers in the line are separated by one space. The last line contains number 0.

Output

The output contains one line for each but the last line of the input. If x is a positive integer contained in a line of the input, the line in the output will contain x - 1 in prime base representation. All numbers in the line are separated by one space. There is no line in the output corresponding to the last null'' line of the input.

Sample Input

17 1

5 1 2 1

509 1 59 1

Sample Output

2 4

3 2

13 1 11 1 7 1 5 1 3 1 2 1

Source

Central Europe 1997

题目类型：素因子分解

算法分析：先将输入的数转换成整数val，然后val自减一，最后将val进行素因子分解就可以了，注意要先打出素数表，否则会TLE

#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

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-6;
const double PI = 2 * acos (0.0);
const int maxn = 1e5 + 66;


LL aa[maxn], maxval;

inline LL power (LL a, LL num)
{
    LL res = 1;
    for (LL i = 1; i <= num; i++)
        res *= a;

    return res;
}


LL primelist[maxn], prime_len;
bool prime[maxn];

void Get ()
{
    memset (prime, true, sizeof (prime));
    prime_len = 0;
    for (LL i = 2; i < maxn - 6; i++)
    {
        if (prime[i])
            primelist[prime_len++] = i;

        for (LL j = 0; j < prime_len; j++)
        {
            if (i * primelist[j] > maxn - 6)
                break;

            prime[i*primelist[j]] = false;
            if (i % primelist[j] == 0)
                break;
        }
    }
}


void PP (LL val)
{
    memset (aa, 0, sizeof (aa));
    maxval = -INF;
    for (LL i = 0; i < prime_len && primelist[i] * primelist[i] <= val; i++)
    {
        while (val % primelist[i] == 0)
        {
            aa[primelist[i]]++;
            maxval = max (maxval, primelist[i]);
            val /= primelist[i];
        }
    }

    if(val > 1)
    {
        aa[val]++;
        maxval = max (maxval, val);
    }
}


int main()
{
    Get ();

//    CPPFF;
    string s;
    while (getline (cin, s))
    {
        if(s == "0")
            break;

        stringstream ss (s);
        LL res = 1, ta, tb;
        while (ss >> ta >> tb)
        {
            res *= power (ta, tb);
        }

        PP (res - 1);
        for (LL i = maxval; i >= 2; i--)
        {
            if (i == maxval && aa[i])
            {
                printf ("%lld %lld", i, aa[i]);
            }

            else if (aa[i])
                printf (" %lld %lld", i, aa[i]);
        }
        puts ("");
    }
    return 0;
}

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

#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

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-6;

const double PI = 2 * acos (0.0);

const int maxn = 1e5 + 66;

LL aa[maxn], maxval;

inline LL power (LL a, LL num)

{

LL res = 1;

for (LL i = 1; i <= num; i++)

res *= a;

return res;

}

LL primelist[maxn], prime_len;

bool prime[maxn];

void Get ()

{

memset (prime, true, sizeof (prime));

prime_len = 0;

for (LL i = 2; i < maxn - 6; i++)

{

if (prime[i])

primelist[prime_len++] = i;

for (LL j = 0; j < prime_len; j++)

{

if (i * primelist[j] > maxn - 6)

break;

prime[i*primelist[j]] = false;

if (i % primelist[j] == 0)

break;

}

void PP (LL val)

{

memset (aa, 0, sizeof (aa));

maxval = -INF;

for (LL i = 0; i < prime_len && primelist[i] * primelist[i] <= val; i++)

{

while (val % primelist[i] == 0)

{

aa[primelist[i]]++;

maxval = max (maxval, primelist[i]);

val /= primelist[i];

}

if(val > 1)

{

aa[val]++;

maxval = max (maxval, val);

}

int main()

{

Get ();

// CPPFF;

string s;

while (getline (cin, s))

{

if(s == "0")

break;

stringstream ss (s);

LL res = 1, ta, tb;

while (ss >> ta >> tb)

{

res *= power (ta, tb);

}

PP (res - 1);

for (LL i = maxval; i >= 2; i--)

{

if (i == maxval && aa[i])

{

printf ("%lld %lld", i, aa[i]);

}

else if (aa[i])

printf (" %lld %lld", i, aa[i]);

}

puts ("");

}

return 0;

}

poj1330

poj1442

超客的水墨烟雨

poj1365

分类目录

标签云

书签

近期评论

2024年12月
一	二	三	四	五	六	日
« 6月
						1
2	3	4	5	6	7	8
9	10	11	12	13	14	15
16	17	18	19	20	21	22
23	24	25	26	27	28	29
30	31