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Test instructions

For a waiting set {AI}, use it to complete the backpack for all the numbers in the 1-m , and can fill a pack of n sizes;

Now give n kinds of backpack {b1...bn} and M, find the minimum size of this set and give a scheme arbitrarily;

Exercises

The simplest set for this problem is {bi}, and if the {BI} collection is illegal, the original problem is not solved;

Judging this set is not legal equivalent to the existence of two elements in the set i,j and I+j? The set (I+J<=M);

Directly done is O (n^2), and can be found to convert the original B array into a generating function, the problem can be optimized with convolution;

A good judgment has a solution, then is to optimize the set, so that the elements inside the least;

The naïve idea is to enumerate the elements from small to large, and then determine whether the element can be represented by an already selected element, or the element will be selected, otherwise not added;

The time complexity of the algorithm is O (n*m), each time to do a complete backpack;

And we can find that it is optional to judge this element is not so troublesome, because given the B array has been the backpack completed, so only need to determine whether the current number can be two i,j, the same is the FFT solution;

Time complexity O (MLOGM);

Code:

#include <math.h> #include <stdio.h> #include <string.h> #include <algorithm> #define N 2097152using namespace Std;const double Pi=acos ( -1.0), struct cp{double x,y;cp () {}CP (double _,double __): X (_), Y (__) {} Friend CP operator + (CP A,CP b) {return CP (A.X+B.X,A.Y+B.Y);} Friend CP operator-(CP A,CP B) {return CP (A.X-B.X,A.Y-B.Y);} Friend CP operator * (CP A,CP B) {return CP (a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}} A[n],b[n];void FFT (CP *a,int len,int type) {int i,j,h,t;for (i=0,t=0;i<len;i++) {if (i>=t) swap (a[i],a[t]); for (j= len>>1; (T^=J) <j;j>>=1);} for (h=2;h<=len;h<<=1) {cp-WN (cos (2*pi*type/h), sin (2*pi*type/h)), for (I=0;I&LT;LEN;I+=H) {CP W (1,0), temp; for (J=0;J&LT;H&GT;&GT;1;J++,W=W*WN) {temp=w*a[i+j+ (h>>1)];a[i+j+ (h>>1)]=a[i+j]-temp;a[i+j]=a[i+j]+ temp;}}} if (type==-1) for (i=0;i<len;i++) A[i].x/=len;} int St[n],top;bool A[n],b[n];int Main () {int n,m,len,i,k;scanf ("%d%d", &n,&m), for (i=1;i<=n;i++) {scanf ("%d") ", &k); a[k].x=1; A[k]=1;} for (i=1<<25;i;i>>=1) if (m&i) {len=i<<2;break;} FFT (a,len,1); for (i=0;i<len;i++) b[i]=a[i]*a[i]; FFT (b,len,-1); for (i=1;i<=m;i++) {if (! a[i]&&b[i].x>1e-5) {puts ("NO"); return 0;} if (a[i]&&b[i].x<1e-5) {st[++top]=i;}} Puts ("YES");p rintf ("%d\n", top), for (i=1;i<=top;i++) printf ("%d", St[i]); return 0;}

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