POJ 2739 Sum of Consecutive Prime Numbers
Sum of Consecutive Prime Numbers
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13050 | Accepted: 7255 |
Description
Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53. The integer 41 has three representations 2+3+5+7+11+13, 11+13+17, and 41. The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime
numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20.
Your mission is to write a program that reports the number of representations for the given positive integer.
numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20.
Your mission is to write a program that reports the number of representations for the given positive integer.
Input
The
input is a sequence of positive integers each in a separate line. The
integers are between 2 and 10 000, inclusive. The end of the input is
indicated by a zero.
Output
The
output should be composed of lines each corresponding to an input line
except the last zero. An output line includes the number of
representations for the input integer as the sum of one or more
consecutive prime numbers. No other characters should be inserted in the
output.
Sample Input
2
3
17
41
20
666
12
53
0
Sample Output
1
1
2
3
0
0
1
2
Source
求连续素数和。用dp就好。dp[x]就是从i开始到x的连续素数和。当dp[x]大于规定的数时就break。
依然是筛法求素数。
1 #include <iostream>
2 #include <cstdlib>
3 #include <algorithm>
4 #include <cstring>
5
6 #define DATASIZE 10005
7 #define PRIMECOUNT 1500
8
9 using namespace std;
10
11 int prime[PRIMECOUNT];
12 int dp[PRIMECOUNT];
13 bool flag[DATASIZE];
14
15 void generate_prime(void)
16 {
17 memset(flag,true,sizeof(bool)*DATASIZE);
18 int current = 0;
19
20 for(int i = 2;i*i <= DATASIZE;++i)
21 if(flag[i])
22 for(int j = 2*i;j <= DATASIZE;j += i)
23 flag[j] = false;
24 for(int i = 2;i != DATASIZE;++i)
25 if(flag[i])
26 prime[current++] = i;
27 }
28
29 int main(void)
30 {
31 generate_prime();
32 int num,count;
33
34 while(cin >> num && num)
35 {
36 count = 0;
37 for(int i = 0;i != PRIMECOUNT;++i)
38 {
39 memset(dp,0,sizeof(int)*PRIMECOUNT);
40 dp[i] = prime[i];
41 if(prime[i] == num)
42 count++;
43 for(int j = i+1;j != PRIMECOUNT;++j)
44 {
45 dp[j] = dp[j-1]+prime[j];
46 if(dp[j] == num)
47 {
48 count++;
49 break;
50 }
51 else if(dp[j] > num)
52 break;
53 }
54 }
55 cout << count << endl;
56 }
57 }
