Educational Codeforces Round 9 E. Thief in a Shop

【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);//headtypedef 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;}