【打表】POJ-2739 Sum of Consecutive Prime Numbers

Sum of Consecutive Prime Numbers

Time Limit: 1000MS Memory Limit: 65536K   

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

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思路：问一个数能被 多少种 若干连续素数之和 表示出来。

一看到连续，就想到起点终点前缀和。

打表即可。

但是要注意一下，0算是前缀和的第一项。

5 = 5 + 0

5 = 2 + 3

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <ctime>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <climits>#include <iostream>#define LL long longusing namespace std;const int INF = 0x3f3f3f3f;/****************************************/const int N = 11111;bool vis[N];int p[2000], ans[N], cnt, n, pre[2000];void get_prime() {    cnt = 0;    vis[1] = 1;    for(int i = 2; i < N; i++) {        if(!vis[i]) p[cnt++] = i;        for(int j = 0; j < cnt && p[j]*i < N; j++) {            vis[p[j]*i] = 1;            if(i % p[j] == 0) break;        }    }}int prefix(){int i;for(i = 1; i < cnt; i++) {pre[i] = p[i-1] + pre[i-1];}return i;}void table(int dot){for(int i = 1; i < dot; i++) {for(int j = 0; j < i; j++) if(pre[i] - pre[j] < N) {ans[pre[i] - pre[j]]++;}}}int main(){#ifdef J_Sure//freopen("000.in", "r", stdin);//freopen(".out", "w", stdout);#endifget_prime();pre[0] = 0;int dot = prefix();table(dot);while(scanf("%d", &n), n) {printf("%d\n", ans[n]);}return 0;}