Complete solution to a problem not understand what he was saying(Must be the solution to a problem because of the collapse of the formula fight was not too bad ...... sibutramine English)
Since \ (C = \ SUM x_i \ Times a_i \) , we can \ (x_i \) to binary decomposition, and then consider each one
Set \ (f_ {i, j} \) represents consideration of the \ (C \) binary section \ (I \) bits and \ (0 \) to \ (i-1 \) bits are already on, carry as \ (j \) the number of programs, then consider all \ (x_i \) present this one, each \ (a_i \) up to choose one with a similar backpack of stuff calculated \ (g_j \) represents the total carry is (J \) \ number of program transfer is \ (F_ {I +. 1, J} + = G_ {2J + C_ {I +. 1}} \) , where \ (c_ {i} \) represents \ ( C \) binary section \ (I \) bits
Time complexity \ (O (can live) \)
//quming
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
const int P=1e9+7;
inline void upd(R int &x,R int y){(x+=y)>=P?x-=P:0;}
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
R int res=1;
for(;y;y>>=1,x=mul(x,x))(y&1)?res=mul(res,x):0;
return res;
}
const int N=1e5+5;
char ss[N];int a[N],f[2][N],g[N],st[N],sum,s,n,T,t,top,bg,ed;
void biu(){
bg=1,ed=strlen(ss+1),top=0;
fp(i,bg,ed)ss[i]-='0';
while(bg<=ed){
R int ret=0,tmp;
fp(i,bg,ed)tmp=ret*10+ss[i],ss[i]=tmp>>1,ret=tmp&1;
st[++top]=ret;
while(bg<=ed&&!ss[bg])++bg;
}
memset(f,0,sizeof(f));
f[t=0][0]=1,sum=0;
fp(i,1,n)sum+=a[i];
fp(k,1,top){
memset(f[t^1],0,(sum+1)<<2);
fp(j,sum+1,sum<<1)f[t][j]=0;
s=-233333;
fd(j,sum,0)if(f[t][j]){s=j;break;}
fp(i,1,n){
fd(j,s,0)upd(f[t][j+a[i]],f[t][j]);
s+=a[i];
}
fp(j,0,sum)f[t^1][j]=f[t][j<<1|st[k]];
t^=1;
}
printf("%d\n",f[t][0]);
}
int main(){
// freopen("testdata.in","r",stdin);
for(scanf("%d",&T);T;--T){
scanf("%d%s",&n,ss+1);
fp(i,1,n)scanf("%d",&a[i]);
sort(a+1,a+1+n);
biu();
}
return 0;
}