BNUOJ 27935 我爱背单词（FFT）

Posted 2022-12-03 AC_Arthur

题目链接：点击打开链接

思路：

该题暴力当然可以过，   如果数据量加大，  我们还有一种nlogn的算法：FFT

仔细观察这个复习单词量的累加方式可以发现， 这是一个卷积， 可以用FFT加速算法。

细节参见代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 4000 + 10;
int T,n,m,b[maxn], a[maxn];
struct Complex 
    double x, y; // 实部和虚部 x + yi
    Complex(double _x = 0.0, double _y = 0.0) 
        x = _x;
        y = _y;
    
    Complex operator +(const Complex &b) const 
        return Complex(x + b.x, y + b.y);
    
    Complex operator -(const Complex &b) const 
        return Complex(x - b.x, y - b.y);
    
    Complex operator *(const Complex &b) const 
        return Complex(x*b.x-y*b.y, x*b.y+y*b.x);
    
;
void change(Complex y[], int len) 
    int i, j, k;
    for(i = 1, j = len/2; i < len-1; i++) 
        if(i < j) swap(y[i], y[j]);
        k = len/2;
        while(j >= k)  j -= k; k /= 2; 
        if(j < k) j += k;
    

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].x /= len;
    

Complex x1[maxn], x2[maxn], ans[maxn];
int main() 
    scanf("%d",&T);
    while(T--) 
        scanf("%d", &n);
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);
        scanf("%d", &m);
        int len1 = n, len2 = 0, len = 1;
        for(int i = 0; i < m; i++) scanf("%d", &b[i]), len2 = max(len2, b[i]);
        while(len < len1 * 2 || len < len2 * 2) len <<= 1;
        for(int i = 0; i < len2; i++) x2[i] = Complex(0, 0);
        for(int i = 0; i < n; i++) x1[i] = Complex(a[i], 0);
        for(int i = 0; i < m; i++) x2[b[i]-1] = Complex(1, 0);
        for(int i = len1; i < len; i++) x1[i] = Complex(0, 0);
        for(int i = len2; i < len; i++) x2[i] = Complex(0, 0);
        fft(x1, len, 1);
        fft(x2, len, 1);
        for(int i = 0; i < len; i++) ans[i] = x1[i] * x2[i];
        fft(ans, len, -1);
        for(int i = 0; i < n; i++) ans[i].x += a[i];
        len = len1 + len2 - 1;
        int q; scanf("%d", &q);
        while(q--) 
            int id;
            scanf("%d", &id);
            if(id > len) printf("0\\n");
            else printf("%d\\n", (int)round(ans[id-1].x));
        
    
    return 0;