2013长沙网络赛 G Goldbach （FFT）

转载请注明出处，谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents    by---cxlove

太逗了。。。比赛的时候，误以为素数会很多。。。。

然后 就想歪 了，开始搞FFT。

其实发现主要 是a + b + c的情况不好处理。

先将a + b的情况FFT一下，然后 再 + c FFT一次。

num[i]表示a + b = i的个数， sum[i] 表示 a + b + c = i的个数。

但是去重比较麻烦，如果 a + a = i的情况最好单独考虑，从num[i]中去掉，查询的时候单独统计。

这样的话和c合并，可能出现a + b + a的情况，这种情况也需要在查询的时候单调搞一下，最后再去掉排列组合的情况。

赛后发现只有8000个素数，所有情况平方都可以预处理

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#define lson step << 1
#define rson step << 1 | 1
#define lowbit(x) (x & (-x))
#define Key_value ch[ch[root][1]][0] 
using namespace std;
typedef long long LL;
const int N = 80005;
const LL MOD = 1000000007;
int prime[N] , flag[N] , cnt = 0;
void init () {
    flag[0] = flag[1] = 1;
    for (int i = 2 ; i < N ; i ++) {
        if (flag[i]) continue;
        prime[cnt ++] = i;
        for (int j = 2 ; j * i < N ; j ++)
            flag[i * j] = 1;
    }
}
//FFT copy from kuangbin    
const double pi = acos (-1.0);    
// Complex  z = a + b * i      
struct Complex {    
    double a, b;    
    Complex(double _a=0.0,double _b=0.0):a(_a),b(_b){}    
    Complex operator + (const Complex &c) const {    
        return Complex(a + c.a , b + c.b);    
    }    
    Complex operator - (const Complex &c) const {    
        return Complex(a - c.a , b - c.b);    
    }    
    Complex operator * (const Complex &c) const {    
        return Complex(a * c.a - b * c.b , a * c.b + b * c.a);    
    }    
};    
//len = 2 ^ k    
void change (Complex y[] , int len) {    
    for (int i = 1 , j = len / 2 ; i < len -1 ; i ++) {    
        if (i < j) swap(y[i] , y[j]);    
        int k = len / 2;    
        while (j >= k) {    
            j -= k;    
            k /= 2;    
        }    
        if(j < k) j += k;    
    }     
}    
// FFT     
// len = 2 ^ k    
// on = 1  DFT    on = -1 IDFT    
void FFT (Complex y[], int len , int on) {    
    change (y , len);    
    for (int h = 2 ; h <= len ; h <<= 1) {    
        Complex wn(cos (-on * 2 * pi / h), sin (-on * 2 * pi / h));    
        for (int j = 0 ; j < len ; j += h) {    
            Complex w(1 , 0);    
            for (int k = j ; k < j + h / 2 ; k ++) {    
                Complex u = y[k];    
                Complex t = w * y [k + h / 2];    
                y[k] = u + t;    
                y[k + h / 2] = u - t;    
                w = w * wn;    
            }    
        }    
    }    
    if (on == -1) {    
        for (int i = 0 ; i < len ; i ++) {    
            y[i].a /= len;    
        }    
    }    
}  
LL sum[N << 2] , num[N << 2];  
Complex x1[N << 2] , x2[N << 2];    
int main (){
    int n = 80000;
    init ();
    int len = n;
    int l = 1;
    while (l < len * 2) l <<= 1;  
    for (int i = 0 ; i <= n ; i ++) {  
        if (flag[i] == 0)  x1[i] = Complex (1 , 0);
        else  x1[i] = Complex (0 , 0);
    }  
    for (int i = n + 1 ; i < l ; i ++)
        x1[i] = Complex (0 , 0);
    FFT(x1 , l , 1);  
    for (int i = 0 ; i < l ; i ++) {  
        x1[i] = x1[i] * x1[i];  
    }  
    FFT(x1 , l , -1); 
    for (int i = 0 ; i <= n ; i ++) {  
        num[i] = (LL)(x1[i].a + 0.5); 
    } 
    for (int i = 0 ; i <= n ; i ++) {
        if (flag[i] == 0) 
            num[i * 2] --;
    } 
    for (int i = 0 ; i <= n ; i ++) {
        num[i] /= 2;
    }

    
    for (int i = 0 ; i <= n ; i ++) {  
        if (flag[i] == 0)  x1[i] = Complex (1 , 0);
        else  x1[i] = Complex (0 , 0);
    }  
    for (int i = n + 1 ; i < l ; i ++)
        x1[i] = Complex (0 , 0);
    for (int i = 0 ; i <= n ; i ++) {
        x2[i] = Complex (num[i] , 0);
    }  
    for (int i = n + 1 ; i < l ; i ++)
        x2[i] = Complex (0 , 0);
    FFT(x1 , l , 1);  
    FFT(x2 , l , 1);
    for (int i = 0 ; i < l ; i ++) {  
        x1[i] = x1[i] * x2[i];  
    }  
    FFT(x1 , l , -1);  
    for (int i = 0 ; i <= n ; i ++) {  
        sum[i] = (LL)(x1[i].a + 0.5); 
    }
    while (scanf ("%d" , &n) != EOF) {    
        LL ans = flag[n] == 0 ? 1 : 0;    // a
        for (int i = 0 ; i < cnt && prime[i] * prime[i] <= n ; i ++) {   // a * b
            if (n % prime[i]) continue;
            if (flag[n / prime[i]] == 0) {
                ans ++;
            }
        }
        for (int i = 0 ; i < cnt && prime[i] <= n ; i ++) {
            for (int j = i ; j < cnt && 1LL * prime[i] * prime[j] <= n ; j ++) {
                if (flag[n - prime[i] * prime[j]] == 0) {
                    ans ++;  // a * b + c
                }
            }
        }
        for (int i = 0 ; i < cnt && prime[i] <= n ; i ++) {
            for (int j = i ; j < cnt && 1LL * prime[i] * prime[j] <= n ; j ++) {
                for (int k = j ; k < cnt && 1LL * prime[i] * prime[j] * prime[k] <= n ; k ++) {
                    if (prime[i] * prime[j] * prime[k] == n)
                        ans ++;   // a * b * c
                }
            }
        }
        int tot ;
        if (n % 2 == 0 && flag[n / 2] == 0) tot = 1;
        else tot = 0;
        ans = (ans + (LL)num[n] + tot) % MOD;   // a + b
        tot = 0;
        for (int i = 0 ; i < cnt ; i ++) {
            if (n - prime[i] * 2 <= 0) break;
            if (flag[n - prime[i] * 2] == 0) tot ++;
        }
        ans = (ans + (sum[n] + 2 * tot) / 3) % MOD;   // a + b + c
        if (n % 3 == 0 && flag[n / 3] == 0) ans ++;
        printf ("%lld\n" , ans);
    }
    return 0;
}