• POJ 3132 Sum of Different Primes ( 满背包问题)


    Sum of Different Primes
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 3280   Accepted: 2040

    Description

    A positive integer may be expressed as a sum of different prime numbers (primes), in one way or another. Given two positive integers n and k, you should count the number of ways to express nas a sum of k different primes. Here, two ways are considered to be the same if they sum up the same set of the primes. For example, 8 can be expressed as 3 + 5 and 5 + 3 but the are not distinguished.

    When n and k are 24 and 3 respectively, the answer is two because there are two sets {2, 3, 19} and {2, 5, 17} whose sums are equal to 24. There are not other sets of three primes that sum up to 24. For n = 24 and k = 2, the answer is three, because there are three sets {5, 19}, {7, 17} and {11, 13}. For n = 2 and k = 1, the answer is one, because there is only one set {2} whose sum is 2. For n = 1 and k = 1, the answer is zero. As 1 is not a prime, you shouldn’t count {1}. For n = 4 and k = 2, the answer is zero, because there are no sets of two different primes whose sums are 4.

    Your job is to write a program that reports the number of such ways for the given n and k.

    Input

    The input is a sequence of datasets followed by a line containing two zeros separated by a space. A dataset is a line containing two positive integers n and k separated by a space. You may assume that n ≤ 1120 and k ≤ 14.

    Output

    The output should be composed of lines, each corresponding to an input dataset. An output line should contain one non-negative integer indicating the number of the ways for n and k specified in the corresponding dataset. You may assume that it is less than 231.

    Sample Input

    24 3 
    24 2 
    2 1 
    1 1 
    4 2 
    18 3 
    17 1 
    17 3 
    17 4 
    100 5 
    1000 10 
    1120 14 
    0 0

    Sample Output

    2 
    3 
    1 
    0 
    0 
    2 
    1 
    0 
    1 
    55 
    200102899 
    2079324314

    Source

    题目大意:

      这道题是说,给你一个数字n,然后让你用仅仅含有k个素数的集合来描述(就是取k个素数使得他们的和是n),问这样的集合有多少个.

    解题思路:

      一看到后,就想到了背包问题,把这个背包塞满的问题,定义如下状态:

      dp[i][j]表示的是把数字i分成j个素数的集合的所有情况数。

      初始状态:dp[1][1] = 0;//从题目中的信息知道的

           dp[0][0] = 1;//这个初值条件是硬加上去的,要不然的话dp[0][0]就无法递推了,永远都是0了。QAQ

    代码:

     1 # include<cstdio>
     2 # include<iostream>
     3 # include<fstream>
     4 # include<algorithm>
     5 # include<functional>
     6 # include<cstring>
     7 # include<string>
     8 # include<cstdlib>
     9 # include<iomanip>
    10 # include<numeric>
    11 # include<cctype>
    12 # include<cmath>
    13 # include<ctime>
    14 # include<queue>
    15 # include<stack>
    16 # include<list>
    17 # include<set>
    18 # include<map>
    19 
    20 using namespace std;
    21 
    22 const double PI=4.0*atan(1.0);
    23 
    24 typedef long long LL;
    25 typedef unsigned long long ULL;
    26 
    27 # define inf 999999999
    28 # define MAX 1120+4
    29 
    30 
    31 int dp[MAX][MAX];//dp[i][j]表示把数字i分解为由j个素数组成的集合数目
    32 int book[MAX];
    33 int prime[MAX];
    34 int len;
    35 
    36 
    37 void init()
    38 {
    39     for ( int i = 2;i <= MAX;i++ )
    40     {
    41         book[i] = 1;
    42     }
    43     for ( int i = 2;i <= MAX;i++ )
    44     {
    45         if ( book[i]==1 )
    46         {
    47             for ( int j = 2*i;j <= MAX;j+=i )
    48             {
    49                 book[j] = 0;
    50             }
    51         }
    52     }
    53     len = 0;
    54 
    55     for ( int i = 2;i <= MAX;i++ )
    56     {
    57         if ( book[i]==1 )
    58         {
    59             prime[++len] = i;
    60         }
    61     }
    62 }
    63 
    64 void solve()
    65 {
    66         memset(dp,0,sizeof(dp));
    67         dp[0][0] = 1;
    68         dp[1][1] = 0;
    69         for ( int i = 1;i <= len;i++ )
    70         {
    71             for ( int j = MAX;j >= prime[i];j-- )
    72             {
    73                 for ( int k = 1;k <= 14;k++ )
    74                 {
    75                     dp[j][k] += dp[j-prime[i]][k-1];
    76                 }
    77             }
    78         }
    79 }
    80 
    81 
    82 int main(void)
    83 {
    84 
    85     init();
    86     int n,k;
    87     while ( cin>>n>>k )
    88     {
    89         if ( n==0&&k==0 )
    90             break;
    91         solve();
    92         cout<<dp[n][k]<<endl;
    93     }
    94 
    95 
    96     return 0;
    97 }
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  • 原文地址:https://www.cnblogs.com/wikioibai/p/4446085.html
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