XTUOJ1250 Super Fast Fourier Transform 暴力

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+;
const int INF=0x3f3f3f3f;
const int mod=1e9+;
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=;i<=n;++i){
        scanf("%d",&a[i]);
        c[i]=a[i];
      }
      sort(a+,a++n);
      int cnt1=unique(a+,a++n)-a-;
      memset(k1,,sizeof(k1));
      for(int i=;i<=n;++i){
        ++k1[lower_bound(a+,a++cnt1,c[i])-a];
      }
      for(int i=;i<=m;++i){
        scanf("%d",&b[i]);
        c[i]=b[i];
      }
      sort(b+,b++m);
      int cnt2=unique(b+,b++m)-b-;
      memset(k2,,sizeof(k2));
      for(int i=;i<=m;++i){
        ++k2[lower_bound(b+,b++cnt2,c[i])-b];
      }
      LL ret=;
      for(int i=;i<=cnt1;++i){
        for(int j=;j<=cnt2;++j){
          LL tmp=sqrt(abs(a[i]-b[j]));
          ret+=1ll*k1[i]*k2[j]*tmp;
        }
      }
      printf("%I64d\n",ret);
    }
    return ;
}