BZOJ 2064: *

最坏的情况是全部合起来然后拆分，n+m-2

每找到一个和相等的子集，答案-2

问题转化为找到最多的子集

转化为随机序列前缀和相等

#include<cstdio>
#include<algorithm>
using namespace std;
int a[15],b[15],Suma[2005],Sumb[2005],F[2005][2005],ID[2005];
int lowbit(int x){
	return x&(-x);
}
int main(){
	int n,m;
	scanf("%d",&n); for (int i=0; i<n; i++) scanf("%d",&a[i]);
	scanf("%d",&m); for (int i=0; i<m; i++) scanf("%d",&b[i]);
	for (int i=0; i<max(n,m); i++) ID[1<<i]=i;
	for (int i=1; i<(1<<n); i++) Suma[i]=Suma[i^lowbit(i)]+a[ID[lowbit(i)]];
	for (int i=1; i<(1<<m); i++) Sumb[i]=Sumb[i^lowbit(i)]+b[ID[lowbit(i)]];
	for (int i=0; i<(1<<n); i++)
		for (int j=0; j<(1<<m); j++){
			if (Suma[i]==Sumb[j]) F[i][j]++;
			for (int k=0; k<n; k++) if (!(i&(1<<k))) F[i|(1<<k)][j]=max(F[i|(1<<k)][j],F[i][j]);
			for (int k=0; k<m; k++) if (!(j&(1<<k))) F[i][j|(1<<k)]=max(F[i][j|(1<<k)],F[i][j]);
		}
	printf("%d
",n+m+2-2*F[(1<<n)-1][(1<<m)-1]);
	return 0;
}