XJTUOJ wmq的A×B Problem FFT/NTT

wmq的A×B Problem

发布时间: 2017年4月9日 17:06   最后更新: 2017年4月9日 17:07   时间限制: 3000ms   内存限制: 512M

描述

这是一个非常简单的问题。

wmq如今开始学习乘法了！他为了训练自己的乘法计算能力，写出了n个整数，并且对每两个数a,b都求出了它们的乘积a×b。现在他想知道，在求出的n(n−1)2个乘积中，除以给定的质数m余数为k(0≤k<m)的有多少个。

输入

第一行为测试数据的组数。

对于每组测试数据，第一行为2个正整数n,m,2≤n,m≤60000，分别表示整数的个数以及除数。

接下来一行有n个整数，满足0≤ai≤109。

保证总输出行数∑m≤3×105。

输出

对每组数据输出m行，其中第i行为除以m余数为(i−1)的有多少个。

样例输入1

2
4 5
2 0 1 7
4 2
2 0 1 6

样例输出1

3
0
2
0
1
6
0

 

题解：

　　m是素数，显然，利用原根性质

　　i -> g^i

　　此时g为m的原根

　　首先 模剩余系 为0~m-1，那么模为 i * j 转化为 g^(i+j)

　　即FFT求解，NTT也可过

　　最后转化回来即可

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 333333+10, M = 1e3+20,inf = 2e9,mod = 469762049;

int MOD;
inline int mul(int a, int b){
    return (long long)a * b % MOD;
}
int power(int a, int b){
    int ret = 1;
    for (int t = a; b; b >>= 1){
        if (b & 1)ret = mul(ret, t);
        t = mul(t, t);
    }
    return ret;
}
int cal_root(int mod)
{
    int factor[20], num = 0, s = mod - 1;
    MOD = mod--;
    for (int i = 2; i * i <= s; i++){
        if (s % i == 0){
            factor[num++] = i;
            while (s % i == 0)s /= i;
        }
    }
    if (s != 1)factor[num++] = s;
    for (int i = 2;; i++){
        int j = 0;
        for (; j < num && power(i, mod / factor[j]) != 1; j++);
        if (j == num)return i;
    }
}
struct Complex {
    long double r , i ;
    Complex () {}
    Complex ( double r , double i ) : r ( r ) , i ( i ) {}
    Complex operator + ( const Complex& t ) const {
        return Complex ( r + t.r , i + t.i ) ;
    }
    Complex operator - ( const Complex& t ) const {
        return Complex ( r - t.r , i - t.i ) ;
    }
    Complex operator * ( const Complex& t ) const {
        return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ;
    }
} ;

void FFT ( Complex y[] , int n , int rev ) {
    for ( int i = 1 , j , t , k ; i < n ; ++ i ) {
        for ( j = 0 , t = i , k = n >> 1 ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ;
        if ( i < j ) swap ( y[i] , y[j] ) ;
    }
    for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) {
        Complex wn = Complex ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w ( 1 , 0 ) , t ;
        for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) {
            for ( int i = k ; i < n ; i += s ) {
                y[i + ds] = y[i] - ( t = w * y[i + ds] ) ;
                y[i] = y[i] + t ;
            }
        }
    }
    if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ;
}

Complex s[N];
int T,n,m,x,num[N];
LL ans[N],mo[N],fmo[N];
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&m);
        int G = cal_root(m);
        for(LL i = 0, t = 1; i < m-1; ++i,t = t*G%m)
            mo[i] = t,fmo[t] = i;
        memset(num,0,sizeof(num));
        LL cnt0 = 0;
        for(int i = 0; i <= m; ++i) ans[i] = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%d",&x);
            x%=m;
            if(x == 0) cnt0++;
            else {
                num[fmo[x]]++;
            }
        }
        int n1 = 1;
        for(n1=1;n1<=(2*m-2);n1<<=1);
        for(int i = 0; i < m; ++i) s[i] = Complex(num[i],0);
        for(int i = m; i < n1; ++i) s[i] = Complex(0,0);
        FFT(s,n1,1);
        for(int i = 0; i < n1; ++i) s[i] = s[i]*s[i];
        FFT(s,n1,-1);
        printf("%lld\n",(LL)cnt0*(n-cnt0)+(LL)cnt0*(cnt0-1)/2);
        for(int i = 0; i <= 2*m-2; ++i) {
            LL now = (LL)(s[i].r+0.5);
            if(i%2==0) now -= num[i/2];
            now/=2;
            ans[mo[i%(m-1)]] += now;
        }
        for(int i = 1; i < m; ++i) printf("%lld\n",ans[i]);
    }
    return 0;
}

 

　　

 

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 533333+10, M = 1e3+20,inf = 2e9;

int MOD;
inline int mul2(int a, int b){
    return (long long)a * b % MOD;
}
int power(int a, int b){
    int ret = 1;
    for (int t = a; b; b >>= 1){
        if (b & 1)ret = mul2(ret, t);
        t = mul2(t, t);
    }
    return ret;
}
int cal_root(int mod)
{
    int factor[26], num = 0, s = mod - 1;
    MOD = mod--;
    for (int i = 2; i * i <= s; i++){
        if (s % i == 0){
            factor[num++] = i;
            while (s % i == 0)s /= i;
        }
    }
    if (s != 1)factor[num++] = s;
    for (int i = 2;; i++){
        int j = 0;
        for (; j < num && power(i, mod / factor[j]) != 1; j++);
        if (j == num)return i;
    }
}

LL P,G;
LL mul(LL x,LL y){
    return (x*y-(LL)(x/(long double)P*y+1e-3)*P+P)%P;
}
LL qpow(LL x,LL k,LL p){
    LL ret=1;
    while(k){
        if(k&1) ret=mul(ret,x);
        k>>=1;
        x=mul(x,x);
    }
    return ret;
}
LL wn[50];
void getwn(){
    for(int i=1; i<=25; ++i){
        int t=1<<i;
        wn[i]=qpow(G,(P-1)/t,P);
    }
}
void NTT_init() {
    P = 3221225473LL,G = 5;
    getwn();
}
int len;
void NTT(LL y[],int op){
    for(int i=1,j=len>>1,k; i<len-1; ++i){
        if(i<j) swap(y[i],y[j]);
        k=len>>1;
        while(j>=k){
            j-=k;
            k>>=1;
        }
        if(j<k) j+=k;
    }
    int id=0;
    for(int h=2; h<=len; h<<=1) {
        ++id;
        for(int i=0; i<len; i+=h){
            LL w=1;
            for(int j=i; j<i+(h>>1); ++j){
                LL u=y[j],t=mul(y[j+h/2],w);
                y[j]=u+t;
                if(y[j]>=P) y[j]-=P;
                y[j+h/2]=u-t+P;
                if(y[j+h/2]>=P) y[j+h/2]-=P;
                w=mul(w,wn[id]);
            }
        }
    }
    if(op==-1){
        for(int i=1; i<len/2; ++i) swap(y[i],y[len-i]);
        LL inv=qpow(len,P-2,P);
        for(int i=0; i<len; ++i) y[i]=mul(y[i],inv);
    }
}
LL s[N];
int T,n,m;
LL ans[N];
int num[N],mo[N],fmo[N],root;
int main() {
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d",&n,&m);
        root = cal_root(m);
        LL cnt0 = 0;
        memset(ans,0,sizeof(ans));
        memset(num,0,sizeof(num));
        for(LL i = 0, t = 1; i < m-1; ++i,t=t*root%m)
            mo[i] = t,fmo[t] = i;
        for(int i = 1; i <= n; ++i) {
            int x;
            scanf("%d",&x);
            x%=m;
            if(x == 0) cnt0++;
            else num[fmo[x]]++;
        }
        for(len = 1; len <= (2*m-2); len<<=1);
        for(int i = 0;i < m; ++i) s[i] = num[i];
        for(int i = m; i < len; ++i) s[i] = 0;
        NTT_init();
        NTT(s,1);
        for(int i = 0; i < len; ++i) s[i] = mul(s[i],s[i]);
        NTT(s,-1);
        for(int i = 0; i <= 2*m-2; ++i) {
            LL now = s[i];
            if(i%2==0) now -= num[i/2];
            now /= 2;
            ans[mo[i%(m-1)]] += now;
        }
        printf("%lld\n",(LL)cnt0*(n-cnt0)+(LL)cnt0*(cnt0-1)/2);
        for(int i = 1; i < m; ++i) printf("%lld\n",ans[i]);
    }
    return 0;
}