此題首先爲利用篩選法求得10000之內的素數,而後對於輸入的每個數字,依次以小於它的連續素數相加,相等則種類數加一,返回,換另外一個素數開始日後繼續相加進行這個過程,最後輸出種類數。
Time Limit: 1 secs, Memory Limit: 32 MBios
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.spa
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.blog
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.three
2 3 17 41 20 666 12 53 0
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() { //篩選法求素數表 vector<int> primes; memset(isprime, true, sizeof(isprime)); isprime[1] = false; for (int i = 2; i <= MAX; i++) {//此處記得爲MAX,而不是sprt(MAX)不然報錯 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; }