分析：因为加起来不超过1e6,所以最多有1000+个不同的数
做法：离散化搞就好了
#include <cstdio> #include <iostream> #include <ctime> #include <vector> #include <cmath> #include <map> #include <queue> #include <algorithm> #include <cstring> using namespace std; typedef long long LL; const int N=1e5+5; const int INF=0x3f3f3f3f; const int mod=1e9+7; int a[N],b[N],c[N],k1[N],k2[N]; int main() { int n,m; while(~scanf("%d%d",&n,&m)){ for(int i=1;i<=n;++i){ scanf("%d",&a[i]); c[i]=a[i]; } sort(a+1,a+1+n); int cnt1=unique(a+1,a+1+n)-a-1; memset(k1,0,sizeof(k1)); for(int i=1;i<=n;++i){ ++k1[lower_bound(a+1,a+1+cnt1,c[i])-a]; } for(int i=1;i<=m;++i){ scanf("%d",&b[i]); c[i]=b[i]; } sort(b+1,b+1+m); int cnt2=unique(b+1,b+1+m)-b-1; memset(k2,0,sizeof(k2)); for(int i=1;i<=m;++i){ ++k2[lower_bound(b+1,b+1+cnt2,c[i])-b]; } LL ret=0; for(int i=1;i<=cnt1;++i){ for(int j=1;j<=cnt2;++j){ LL tmp=sqrt(abs(a[i]-b[j])); ret+=1ll*k1[i]*k2[j]*tmp; } } printf("%I64d ",ret); } return 0; }