CodeForces 632E Thief in a Shop（FFT）

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题意：n种物品每种物品有无限个，每个物品有一个价格，现在问选取k个的所以可能总价。

思路：FFT。如果只选择两件商品的话，很容易想到FFT，对于其他的k，只要类似快速幂来求即可，这样做时间复杂度为O(W*logW*logK)。

其实还可以进一步优化，每一步fft的长度只需要设置为当前的可能最大值就行了，也就是说fft的length也是倍增的，所以复杂度降为了O(W*min(log1000,logK))。

要注意的是因为都变了可能爆精度，所以每一步当某一个值大于1，就把这一位重新设置为1，这样可以确保不会爆。

这一道题还可以用DP来做，具体见 CodeForces 632E Thief in a Shop（DP）。

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<vector>#include<map>#include<queue>#include<stack>#include<string>#include<map>#include<set>#include<ctime>#define eps 1e-6#define LL long long#define pii pair<int, int>#define pb push_back#define mp make_pair//#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;const int MAXN = 2010;//const int INF = 0x3f3f3f3f;const double PI = acos(-1.0);int n, k, a[MAXN], len;//复数结构体struct complex{    double r,i;    complex(double _r = 0.0,double _i = 0.0)    {        r = _r; i = _i;    }    complex operator +(const complex &b)    {        return complex(r+b.r,i+b.i);    }    complex operator -(const complex &b)    {        return complex(r-b.r,i-b.i);    }    complex operator *(const complex &b)    {        return complex(r*b.r-i*b.i,r*b.i+i*b.r);    }};/* * 进行FFT和IFFT前的反转变换。 * 位置i和 （i二进制反转后位置）互换 * len必须取2的幂 */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]);        //交换互为小标反转的元素，i<j保证交换一次        //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的        k = len/2;        while( j >= k)        {            j -= k;            k /= 2;        }        if(j < k) j += k;    }}/* * 做FFT * len必须为2^k形式， * on==1时是DFT，on==-1时是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].r /= len;}complex x1[MAXN*MAXN], x2[MAXN*MAXN], x3[MAXN*MAXN];void solve(int m) {if (m == 1)return;solve(m/2);while (len <= 1000*m)len <<= 1;fft(x1, len, 1);fft(x2, len, 1);for (int i = 0; i < len; i++) {        x1[i] = x1[i] * x2[i];}fft(x1, len, -1);for (int i = 0; i < len; i++)if (x1[i].r > 0.5) x1[i] = x2[i] = complex(1, 0);else  x1[i] = x2[i] = complex(0, 0);if (m&1) {fft(x1, len, 1);fft(x3, len, 1);for (int i = 0; i < len; i++) {x1[i] = x1[i] * x3[i];}fft(x1, len, -1);fft(x3, len, -1);for (int i = 0; i < len; i++)if (x1[i].r > 0.5)x1[i] = x2[i] = complex(1, 0);else x1[i] = x2[i] = complex(0, 0);for (int i = 0; i < len; i++)if (x3[i].r > 0.5)x3[i] = complex(1, 0);else x3[i] = complex(0, 0);}} int main(){freopen("input.txt", "r", stdin);scanf("%d%d", &n, &k);len = 1024;for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);x1[a[i]] = x2[a[i]] = x3[a[i]] = complex(1, 0);}solve(k);for (int i = 0; i <= 1000*k; i++) {if (x1[i].r >= 0.5)printf("%d ", i);}return 0;}

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