【POJ 2739】 Sum of Consecutive Prime Numbers 前綴和預處理 打表 線性篩

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
1
2

題意：給一個數，問它能被分解成幾個連續的素數

思路（預處理+前綴和）：

寫完看有些博主用尺取法，優化的很好。不過這個題數據量小，10000以內的素數才1000個左右，所以可以考慮O（n²）暴力。
1.先線性篩法把素數全部記錄在primes數組裏面
2.然後對該數組求求綴和記錄在sum數組中
3.然後對sum數組枚舉區間左右端點，看(sum[i]-sum[j-1]==n)是否成立即可

AC代碼：

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <math.h>
#include <queue>
#include <stack>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define maxn 10000+500
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
const int maxm = 100000+5;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') ch = getchar();while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x; }
bool vis[maxn];
ll primes[maxn];int tot = 1;
int n;
void Prime()
{
    memset(vis,0,sizeof(vis));
    vis[2] = 0, vis[4] = 1;
    primes[0] = 2;
    for(int i=3;i<maxn;i+=2)
    {
        if(vis[i]==0) primes[tot++] = i;
        for(int j=0;j<tot&&i*primes[j]<maxn;j++)
        {
            vis[i*primes[j]] = 1;
            if(i%primes[j]==0) break;
        }
    }
}
int sum[maxn];
int main()
{
    Prime();
    //cout<<tot<<endl;
    sum[0] = 0;
    for(int i=0;i<tot;i++)
    sum[i+1] = sum[i] + primes[i];
    while(~scanf("%d",&n)&&n)
    {
         int idx = lower_bound(primes,primes+tot,n)-primes;
         if(primes[idx]!=n) idx--;
         int cnt = 0;
         for(int i=idx+1;i>=1;i--)
         {
             for(int j=i;j>=1;j--)
             {
                 if(sum[i]-sum[j-1]==n)  cnt++;
             }
         }
         printf("%d\n",cnt);
    }
    return 0;
}