#include <bits/stdc++.h> using namespace std; #define fru(j,n) for(int j=0; j<(n); ++j) #define tr(it,v) for(auto it=(v).begin(); it!=(v).end(); ++it) #define x first #define y second #define pb push_back #define ALL(G) (G).begin(),(G).end() #if 0 #define DEB printf #else #define DEB(...) #endif typedef long long ll; typedef long long LL; typedef double D; typedef pair<int,int> pii; typedef vector<int> vi; const int inft = 1000000009; const int mod = 1000000007; const int MAXN = 200006,T=1024*1024; ll t[MAXN],pt[MAXN]; ll C[2*T]; // ile aktualnie czeka ll W[2*T]; // wyniki ll W2[2*T]; // wyniki void add(ll* tab,int a,int b,ll x){ a+=T;b+=T; while(a<=b){ if(a%2==1){tab[a]+=x;a++;} if(b%2==0){tab[b]+=x;b--;} a/=2;b/=2; } } ll get(ll *tab, int a){ a+=T; ll ret=0; while(a){ ret+=tab[a]; a/=2; } return ret; } vector<pii> Q; vector<pii> X; bool waitt(int poz,int i){ //czy i-te zgloszenie czeka int wa=get(C,poz); DEB("czas piecz %d(val%d) czeka %d ##",poz,X[poz].x,wa); ll tim=1LL*(wa+1)*X[poz].x+(i-wa>0?t[i-wa-1]:0); DEB("lacznie %lld, poowinno %lld\n",tim,t[i]); return tim>t[i]; } int main() { int n,m; scanf("%d%d",&n,&m); fru(i,n)scanf("%lld",&t[i]); pt[0]=0; fru(i,n)pt[i+1]=pt[i]+t[i]; X.resize(m); fru(i,m){ scanf("%d",&X[i].x); X[i].y=i; } sort(ALL(X)); Q.push_back(pii(m,0)); t[n]=1LL<<50; fru(i,n+1){ //find first that will wait: int p=0,k=m; if(waitt(0,i))k=0; while(k-p>1){ int mid=(p+k)/2; if(waitt(mid,i))k=mid; else p=mid; } DEB("dodaje od %d(val%d)\n",k,X[k].x); add(C,k,m,1); int sum=0; while(Q.back().x<k){ sum+=Q.back().y; int A=Q.back().x; Q.pop_back(); int B=min(k,Q.back().x); // [A,B) dodaj do wyniku, kazdy czeka z sum DEB("zeruje przedzial [%d,%d) z %d\n",A,B,sum); add(C,A,B-1,-1LL*sum);//C -sumy na przedziale add(W,A,B-1,1LL*sum*(sum+1)/2); ll dt=pt[i]-pt[i-sum]-1LL*sum*(i-sum>0?t[i-sum-1]:0); DEB("czekaja %lld\n",dt); add(W2,A,B-1,dt); } sum++; if(Q.back().x==k)Q.back().y+=sum; else Q.push_back(pii(k,sum)); } vector<ll> ANS(m); fru(i,m){ ANS[X[i].y]=get(W,i)*X[i].x-get(W2,i); } fru(i,m)printf("%lld\n",ANS[i]); return 0; }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 | #include <bits/stdc++.h> using namespace std; #define fru(j,n) for(int j=0; j<(n); ++j) #define tr(it,v) for(auto it=(v).begin(); it!=(v).end(); ++it) #define x first #define y second #define pb push_back #define ALL(G) (G).begin(),(G).end() #if 0 #define DEB printf #else #define DEB(...) #endif typedef long long ll; typedef long long LL; typedef double D; typedef pair<int,int> pii; typedef vector<int> vi; const int inft = 1000000009; const int mod = 1000000007; const int MAXN = 200006,T=1024*1024; ll t[MAXN],pt[MAXN]; ll C[2*T]; // ile aktualnie czeka ll W[2*T]; // wyniki ll W2[2*T]; // wyniki void add(ll* tab,int a,int b,ll x){ a+=T;b+=T; while(a<=b){ if(a%2==1){tab[a]+=x;a++;} if(b%2==0){tab[b]+=x;b--;} a/=2;b/=2; } } ll get(ll *tab, int a){ a+=T; ll ret=0; while(a){ ret+=tab[a]; a/=2; } return ret; } vector<pii> Q; vector<pii> X; bool waitt(int poz,int i){ //czy i-te zgloszenie czeka int wa=get(C,poz); DEB("czas piecz %d(val%d) czeka %d ##",poz,X[poz].x,wa); ll tim=1LL*(wa+1)*X[poz].x+(i-wa>0?t[i-wa-1]:0); DEB("lacznie %lld, poowinno %lld\n",tim,t[i]); return tim>t[i]; } int main() { int n,m; scanf("%d%d",&n,&m); fru(i,n)scanf("%lld",&t[i]); pt[0]=0; fru(i,n)pt[i+1]=pt[i]+t[i]; X.resize(m); fru(i,m){ scanf("%d",&X[i].x); X[i].y=i; } sort(ALL(X)); Q.push_back(pii(m,0)); t[n]=1LL<<50; fru(i,n+1){ //find first that will wait: int p=0,k=m; if(waitt(0,i))k=0; while(k-p>1){ int mid=(p+k)/2; if(waitt(mid,i))k=mid; else p=mid; } DEB("dodaje od %d(val%d)\n",k,X[k].x); add(C,k,m,1); int sum=0; while(Q.back().x<k){ sum+=Q.back().y; int A=Q.back().x; Q.pop_back(); int B=min(k,Q.back().x); // [A,B) dodaj do wyniku, kazdy czeka z sum DEB("zeruje przedzial [%d,%d) z %d\n",A,B,sum); add(C,A,B-1,-1LL*sum);//C -sumy na przedziale add(W,A,B-1,1LL*sum*(sum+1)/2); ll dt=pt[i]-pt[i-sum]-1LL*sum*(i-sum>0?t[i-sum-1]:0); DEB("czekaja %lld\n",dt); add(W2,A,B-1,dt); } sum++; if(Q.back().x==k)Q.back().y+=sum; else Q.push_back(pii(k,sum)); } vector<ll> ANS(m); fru(i,m){ ANS[X[i].y]=get(W,i)*X[i].x-get(W2,i); } fru(i,m)printf("%lld\n",ANS[i]); return 0; } |