題目

Source

http://codeforces.com/contest/632/problem/E

Description

A thief made his way to a shop.

As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product of kind i is ai.

The thief is greedy, so he will take exactly k products (it's possible for some kinds to take several products of that kind).

Find all the possible total costs of products the thief can nick into his knapsack.

Input

The first line contains two integers n and k (1?≤?n,?k?≤?1000) — the number of kinds of products and the number of products the thief will take.

The second line contains n integers ai (1?≤?ai?≤?1000) — the costs of products for kinds from 1 to n.

Output

Print the only line with all the possible total costs of stolen products, separated by a space. The numbers should be printed in the ascending order.

Sample Input

3 2
1 2 3

5 5
1 1 1 1 1

3 3
3 5 11

Sample Output

2 3 4 5 6
5
9 11 13 15 17 19 21 25 27 33

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分析

題目大概說給有n種價值各一的物品，每種數量都無限多，問取出k個物品能取出的物品價值和的所有情況。

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用母函數解，價值為指數、存不存在為系數，構造多項式求k次冪即可。
這自然想到FFT+快速冪求，這樣時間復雜度才夠。

FFT直接求的話結果的系數最大到達1000¹⁰⁰⁰太爆炸了，當然也可以求一次卷積后非0指數重新賦值成1；不過我想著開頭一次DFT結尾一次IDFT這樣更快、更輕松點，所以用NTT了。。

我NTT模數取1004535809 WA在20，取998244353 WA在21。。看樣子是系數取模后變為0了，數據叼叼的。。于是我就兩個模數都取，然后4000多ms險過了。。

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代碼

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define MAXN 1048576//const long long P=50000000001507329LL; // 190734863287 * 2 ^ 18 + 1 long long P=1004535809; // 479 * 2 ^ 21 + 1 //const long long P=998244353; // 119 * 2 ^ 23 + 1 const int G=3;long long mul(long long x,long long y){return (x*y-(long long)(x/(long double)P*y+1e-3)*P+P)%P; } long long qpow(long long x,long long k,long long p){long long ret=1;while(k){if(k&1) ret=mul(ret,x);k>>=1;x=mul(x,x);}return ret; }long long wn[25]; void getwn(){for(int i=1; i<=21; ++i){int t=1<<i;wn[i]=qpow(G,(P-1)/t,P);} }int len; void NTT(long long 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){long long w=1;for(int j=i; j<i+(h>>1); ++j){long long 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]);long long inv=qpow(len,P-2,P);for(int i=0; i<len; ++i) y[i]=mul(y[i],inv);} } void Convolution(long long A[],long long B[],int n){for(len=1; len<(n<<1); len<<=1);for(int i=n; i<len; ++i){A[i]=B[i]=0;}NTT(A,1); NTT(B,1);for(int i=0; i<len; ++i){A[i]=mul(A[i],B[i]);}NTT(A,-1); }long long A[MAXN],B[MAXN],C[MAXN]; long long cnt[MAXN];int main(){getwn();int n,k,a;scanf("%d%d",&n,&k);int mx=0;for(int i=0; i<n; ++i){scanf("%d",&a);++cnt[a];mx=max(mx,a);}for(len=1; len<mx*k; len<<=1);memcpy(A,cnt,sizeof(cnt));NTT(A,1);memcpy(B,A,sizeof(B));--k;int tmp=k;while(k){if(k&1){for(int i=0; i<len; ++i) B[i]=mul(A[i],B[i]);}for(int i=0; i<len; ++i) A[i]=mul(A[i],A[i]);k>>=1;}NTT(B,-1);P=998244353;getwn();memcpy(A,cnt,sizeof(cnt));NTT(A,1);memcpy(C,A,sizeof(C));k=tmp;while(k){if(k&1){for(int i=0; i<len; ++i) C[i]=mul(A[i],C[i]);}for(int i=0; i<len; ++i) A[i]=mul(A[i],A[i]);k>>=1;}NTT(C,-1);for(int i=0; i<len; ++i){if(B[i] || C[i]) printf("%d ",i);}return 0; }

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轉載于:https://www.cnblogs.com/WABoss/p/5915285.html

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