Educational Codeforces Round 9 E. Thief in a Shop

最新推荐文章于 2021-07-25 13:28:15 发布

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本文介绍了一种解决特定背包问题的方法，通过使用快速傅立叶变换（FFT）来计算不同物品价值组合的可能性。该方法利用布尔数组减少精度误差，并详细展示了如何通过卷积计算所有可能的价值组合。

【题目链接】点击打开链接

【题意】给定N, K, N种物品每种的价值为Ai，必须装满K个物品的背包，求所有能装的价值，从小到大输出。

【解题方法】其实就是长度为1000的价值向量的k次幂，存在该价值就为1，否则为0。然后用FFT求K维卷积，用bool数组可以降低精度误差，同时长度不要直接设为1e6。

【复杂度分析】 O(W*logW*logK)

【AC代码】

//
//Created by BLUEBUFF 2016/1/10
//Copyright (c) 2016 BLUEBUFF.All Rights Reserved
//

#pragma comment(linker,"/STACK:102400000,102400000")
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <time.h>
#include <cstdlib>
#include <cstring>
#include <complex>
#include <sstream> //isstringstream
#include <iostream>
#include <algorithm>
using namespace std;
//using namespace __gnu_pbds;
typedef long long LL;
typedef pair<int, LL> pp;
#define REP1(i, a, b) for(int i = a; i < b; i++)
#define REP2(i, a, b) for(int i = a; i <= b; i++)
#define REP3(i, a, b) for(int i = a; i >= b; i--)
#define CLR(a, b)     memset(a, b, sizeof(a))
#define MP(x, y)      make_pair(x,y)
template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }
const int maxn = (1<<21) + 10;
const int maxm = 100010;
const int maxs = 10;
const int maxp = 1e3 + 10;
const int INF  = 1e9;
const int UNF  = -1e9;
const int mod  = 1e9 + 7;
const double PI = acos(-1);
//head

typedef complex <double> Complex;

void rader(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;
    }
}
void fft(Complex *y, int len, int op) {
    rader(y, len);
    for(int h = 2; h <= len; h <<= 1) {
        double ang = op * 2 * PI / h;
        Complex wn(cos(ang), sin(ang));
        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(op == -1) for(int i = 0; i < len; i++) y[i] /= len;
}
int n, m, k;
Complex x1[maxn], x2[maxn];
bool p[maxn], q[maxn];

void multiply(bool *p, int &n, bool *q, int m){
    int len = 1;
    while(len <= n + m) len <<= 1;
    for(int i = 0; i < len; i++) x1[i] = Complex(i <= n ? p[i] : 0, 0);
    for(int i = 0; i < len; i++) x2[i] = Complex(i <= m ? q[i] : 0, 0);
    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 <= n + m; i++) p[i] = x1[i].real() > 0.5;
    n += m;
}

int main()
{
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++){
        int x;
        scanf("%d", &x);
        q[x] = 1;
    }
    m = 1000;
    n = 0;
    p[0] = 1;
    while(k)
    {
        if(k & 1) multiply(p, n, q, m);
        if(k > 1) multiply(q, m, q, m);
        k >>= 1;
    }
    for(int i = 1; i <= n; i++) if(p[i]) printf("%d ", i); printf("\n");
    return 0;
}