Acdream1174 merge similar items

The question is:

Given n, a [1]... A [n], and 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 [] Quickly calculates the Ans 3 seconds. When this multiplication table is drawn, the B [I] is merged, then, after merging the same class items, the efficiency can be achieved. mlogn cannot use the binary of STL, and the card constant is too unreasonable.

#include <iostream>#include <cstdio>#include <string.h>#include <algorithm>#include <vector>using namespace std;typedef long long ll;const int max_n =100005;struct point{    int v,num;    point(ll a=0, ll b=0){      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=0, R =n;      while(L<R){        int mid = (L+R)/2;        if(K[mid]<=v) L=mid+1;        else R=mid;      }      return L;}int main(){     ll two=2,one =1;     int cas=0;      /*freopen("data.in","r",stdin);        freopen("data.out","w",stdout);    */while(scanf("%d%d",&n,&m)==2){          int V;          ll num=one*(n-1)*n/two;         for(int i=0; i<n; ++i){              scanf("%d",&V);              A[i]=V;              P[i]=point(V,i);         }         for(int i=0; i<m; ++i)            scanf("%d",&B[i]);         ll colu=0,per=0;         for(int i =0; i<n; i++){             colu=colu+one*A[i]*i;             per=per+A[i];         }         ll ans=0;         for(int i=0; i<m; ++i){             ans = ans+ colu;             colu= colu - per;             ans = ans - one*B[i]*num;             num = num - n;         }         sort(P,P+n);         perAnum[0]=P[0].num;         perAsum[0]=one*P[0].v*P[0].num;         perAper[0]=P[0].v;         K[0]=P[0].v;         for(int i=1; i<n; ++i){             perAnum[i] = perAnum[i-1]+P[i].num;             perAsum[i] = perAsum[i-1]+one*P[i].num*P[i].v;             perAper[i] = perAper[i-1]+P[i].v;             K[i]=P[i].v;         }         for(int i=0; i<m; ++i){             int loc = binser(n,B[i]);             if(loc<=0)continue;             loc-=1;             ll perA = perAsum[loc]-perAper[loc]*i;             ll perB = B[i]*(perAnum[loc]-one*(loc+1)*i);             ans = ans - ( perA-perB )*two;         }          printf("%I64d\n",ans);    }    return 0;}

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Acdream1174 merge similar items