POJ 2739 - Sum of Consecutive Prime Numbers(素数筛法+前缀和 / 尺取法)

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

2317412066612530

Sample Output

11230012

                                                                   

Way1：

素性测试O（根号n）

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;int num[1500];bool test(int x){    for(int i = 2; i * i <= x; ++i) {        if(x % i == 0) return false;    }    return true;}int job(){    int k= 1;    num[0] = 0;    num[k] = 2;    for(int i = 3; i < 10000; ++i) {        if(test(i)) {            k++;            num[k] = num[k - 1] + i;        }    }    return k;}int main(){    int n;    int s = job();    while(~scanf("%d", &n)) {        if(n == 0) break;        int ans = 0;        int tmp = 0;        for(int i = 0; i < s; ++i) {            for(int j = i + 1; j < s; ++j) {                if(num[j] - num[i] > n) break;                if(num[j] - num[i] == n) ans++;            }        }        printf("%d\n", ans);    }    return 0;}

Way2：

埃氏素数筛法O（nloglogn）

CODE:

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAX = 10005;int pri[MAX], vis[MAX], ans[MAX];int main(){    int n, p = 0;    memset(ans, 0, sizeof(ans));    memset(vis, 1, sizeof(vis));    for(int i = 2; i <= MAX; ++i) {        if(!vis[i]) continue;        pri[p++] = i;        for(int j = 2*i; j < MAX; j+=i) vis[j] = 0;    }    for(int i = 0; i < p; ++i) {        int s = 0;        for(int j = i; j < p; ++j) {            s += pri[j];            if(s > MAX) break;            ans[s]++;        }    }    while(~scanf("%d", &n)) {        if(n == 0) break;        printf("%d\n", ans[n]);    }    return 0;}

尺取法.

AC.

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAX = 100005;int pri[MAX], vis[MAX];int n, p;void ready(){    p = 0;    memset(vis, 0, sizeof(vis));    for(int i = 2; i < MAX; ++i) {        if(vis[i]) continue;        pri[p++] = i;        for(int j = i*2; j < MAX; j += i) vis[j] = 1;    }}void solve(){    int sum = 0, k = 0, t = 0;    int ans = 0;    while(1) {        while(t <= p && sum < n) sum += pri[t++];        if(pri[t-1] > n) break;        if(sum == n) ans++;        sum -= pri[k++];    }    printf("%d\n", ans);}int main(){//freopen("in", "r", stdin);    ready();    while(~scanf("%d", &n)) {        if(n == 0) break;        solve();    }    return 0;}