acdream1174 合并同类项

这题说的是

给出N,a[1]... a[N]，还有M,b[1]... b[M]
long long ans = 0;
for(int i = 1; i <= N; i ++)
    for(int j = 1; j <= M; j ++)
        ans += abs(a[i] - b[j]) * (i - j); NM【1,50000】 快熟计算出ans 3 秒 当画出这个乘法表后救就会发现 将b[i] 合并，然后合并完同类项就可以做了 效率 mlogn 使用stl的二分还不行卡常数太不合理了

#include <iostream>
#include <cstdio>
#include <string.h>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
const int max_n =;
struct point{
    int v,num;
    point(ll a=, ll b=){
      v=a; num=b;
    }
    bool operator <(const point A)const {
      return v<A.v||(v==A.v&&num<A.num);
    }
}P[max_n];
int n,m;
int B[max_n],A[max_n],K[max_n];
ll perAsum[max_n],perAnum[max_n],perAper[max_n];
int binser(int n, int v){
      int L=, R =n;
      while(L<R){
        int mid = (L+R)/;
        if(K[mid]<=v) L=mid+;
        else R=mid;
      }
      return L;
}
int main()
{
     ll two=,one =;
     int cas=;
      /*freopen("data.in","r",stdin);
        freopen("data.out","w",stdout);
    */while(scanf("%d%d",&n,&m)==){
          int V;
          ll num=one*(n-)*n/two;
         for(int i=; i<n; ++i){
              scanf("%d",&V);
              A[i]=V;
              P[i]=point(V,i);
         }
         for(int i=; i<m; ++i)
            scanf("%d",&B[i]);
         ll colu=,per=;
         for(int i =; i<n; i++){
             colu=colu+one*A[i]*i;
             per=per+A[i];
         }
         ll ans=;
         for(int i=; i<m; ++i){
             ans = ans+ colu;
             colu= colu - per;
             ans = ans - one*B[i]*num;
             num = num - n;
         }
         sort(P,P+n);
         perAnum[]=P[].num;
         perAsum[]=one*P[].v*P[].num;
         perAper[]=P[].v;
         K[]=P[].v;
         for(int i=; i<n; ++i){
             perAnum[i] = perAnum[i-]+P[i].num;
             perAsum[i] = perAsum[i-]+one*P[i].num*P[i].v;
             perAper[i] = perAper[i-]+P[i].v;
             K[i]=P[i].v;
         }
         for(int i=; i<m; ++i){
             int loc = binser(n,B[i]);
             if(loc<=)continue;
             loc-=;
             ll perA = perAsum[loc]-perAper[loc]*i;
             ll perB = B[i]*(perAnum[loc]-one*(loc+)*i);
             ans = ans - ( perA-perB )*two;
         }
          printf("%I64d\n",ans);
    }
    return ;
}