此题首先为利用筛选法求得10000以内的素数，然后对于输入的每一个数字，依次以小于它的连续素数相加，相等则种类数加一，返回，换另一个素数开始往后继续相加进行这个过程，最后输出种类数。
1259. Sum of Consecutive Primes
Constraints
Time Limit: 1 secs, Memory Limit: 32 MB
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.
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
#include <iostream>
#include <cmath>
#include <cstring>
#include <vector>>
using namespace std;
const int MAX = 10000;
bool isprime[10000];
int main() {
//筛选法求素数表
vector<int> primes;
memset(isprime, true, sizeof(isprime));
isprime[1] = false;
for (int i = 2; i <= MAX; i++) {//此处记得为MAX，而不是sprt(MAX)否则报错
if (isprime[i]) {
primes.push_back(i);
int j = i * 2;
while (j <= MAX) {
isprime[j] = false;
j += i;
}
}
}
int number;
while (cin >> number && number != 0) {
int sum = 0;
int count = 0;
for (int i = 0; i < primes.size() && primes[i] <= number; i++) {
for (int j = i; j < primes.size() && primes[j] <= number; j++) {
sum += primes[j];
if (sum == number) {
count++;
sum = 0;
break;
} else if (sum > number) {
sum = 0;
break;
}
}
}
cout << count << endl;
}
return 0;
}