Codeforces Round #251 (Div. 2) D binary

Codeforces Round #251 (Div. 2) D binary

 

It is a good question. First of all, it is not difficult to draw a few more questions. To meet the question conditions, it is possible that the minimum value of array a is increasing, or the maximum value of array B is decreasing, in the beginning, the priority queue was used directly, and the error was found. Because there were not two situations at a time, either a plus or B minus, it was difficult to draw a few more pictures, we need to find that all elements in a with the value of x are greater than or equal to x, and all elements in B are less than or equal to x, so we only need to find this x, according to a few more draws, it is found that x is one of the elements in a and B. Therefore, if multiple paintings are correct, it is directly searched in the array a and B, then, perform a binary search for the answer, and then obtain the final smallest answer. You need to maintain the prefix of a and the suffix of B.

 

 

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              #define ll long long#define eps 1e-8const int inf = 0xfffffff;const ll INF = 1ll<<61;using namespace std;//vector
              
                > G;//typedef pair
               
                 P;//vector
                
                  > ::iterator iter;////map
                 
                  mp;//map
                  
                   ::iterator p;int n,m;ll ans;ll aa[100000 + 55],bb[100000 + 55];ll sa[100000 + 55],sb[100000 + 55];vector
                   
                     G;void init() {memset(aa,0,sizeof(aa));memset(bb,0,sizeof(bb));memset(sa,0,sizeof(sa));memset(sb,0,sizeof(sb));G.clear();}bool input() {while(scanf(%d %d,&n,&m) == 2) {for(int i=1;i<=n;i++) {scanf(%I64d,&aa[i]);G.push_back(aa[i]);}for(int i=1;i<=m;i++) {scanf(%I64d,&bb[i]);G.push_back(bb[i]);}return false;}return true;}ll find(ll x) {ll sum = 0;int pos = lower_bound(aa + 1,aa + n + 1,x) - aa;sum += (pos - 1) * x - sa[pos - 1];pos = lower_bound(bb + 1,bb + m + 1,x) - bb;sum += sb[pos] - (m - pos + 1) * x;return sum;}void cal() {sort(aa + 1,aa + n + 1);sort(bb + 1,bb + m + 1);sort(G.begin(),G.end());for(int i=1;i<=n;i++)sa[i] = sa[i - 1] + aa[i];for(int i=m;i>0;i--)sb[i] = sb[i + 1] + bb[i];ans = INF;for(int i=0;i