【bzoj3441】乌鸦喝水

Source

bzoj3441

Hint

请先思考后再展开

按被删除的顺序考虑每个点，然后按照题意模拟

Solution

请先思考后再展开

被删除的顺序一定是按照【能被操作的次数】为第一关键字，位置作为第二关键字
然后你维护当前轮数和当前走到什么位置（我存了具体位置，其实应该存排名更好写……）
用个树状数组资瓷排名和位置之间的转化即可
时间复杂度为 $O(nlogn)$

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#include<ctime>#include<cstdio>#include<cstring>#include<cstdlib>#include<map>#include<set>#include<queue>#include<deque>#include<stack>#include<bitset>#include<vector>#include<algorithm>#include<iostream>#include<deque>using namespace std;namespace mine{ typedef long long ll; #define double long double const int INF=0x3f3f3f3f; ll () { ll ans=0;char c=getchar();int f=1; while(c<'0' or c>'9') {if(c=='-') f=-1;c=getchar();} while('0'<=c and c<='9') ans=ans*10+c-'0',c=getchar(); return ans*f; } void write(ll num) { if(num<0) {num=-num;putchar('-');} if(num>9) write(num/10); putchar('0'+num%10); } void writeln(int num){write(num);puts("");} #define FR first #define SE second #define MP make_pair #define pr pair<ll,ll> #define PB push_back inline void chmax(ll &x,ll y) {x=x>y?x:y;} inline void chmin(ll &x,ll y) {x=x<y?x:y;} const int MAX_N=1e5+10; int bin[30]; struct BIT { int bit[MAX_N];BIT(){memset(bit,0,sizeof bit);} int lowbit(int x) {return x&-x;} void add(int x,int c) {while(x<MAX_N) bit[x]+=c,x+=lowbit(x);} int ask(int x) {int ans=0;while(x>=1) ans+=bit[x],x-=lowbit(x);return ans;} int findk(int k) { int ans=0; for(int i=20;i>=0;i--) if(ans+bin[i]<MAX_N and k>=bit[ans+bin[i]]) k-=bit[ans+bin[i]],ans+=bin[i]; return ans; } }bit; pr a[MAX_N]; void main() { bin[0]=1;for(int i=1;i<30;i++) bin[i]=bin[i-1]<<1; int n,m,all;scanf("%d%d%d",&n,&m,&all); for(int i=1;i<=n;i++) a[i].FR=all-qread()+1; for(int i=1;i<=n;i++) a[i].FR=ceil((double)a[i].FR/qread()),a[i].SE=i,bit.add(i,1); sort(a+1,a+n+1); ll ans=0,dec=0;int turn=1,pos=1;//pos=应从哪里开始 for(int i=1;i<=n and turn<=m;i++) { int rk=bit.ask(pos); ll tmp=a[i].FR-dec;int me=bit.ask(a[i].SE); if(tmp>0) { int rk2=(tmp+rk-1)%(n-i+1),t2=(tmp+rk-1)/(n-i+1); if(rk2==0) rk2=n-i+1,t2--; if(turn+t2>m) {ans+=ll(m-turn)*(n-i+1)+(n-i+1-rk+1);break;} turn+=t2;dec+=tmp;ans+=tmp; if((me==n-i+1 and rk2==n-i) or rk2==n-i+1) rk2=1,turn++; else if(rk2<me) rk2++; bit.add(a[i].SE,-1); pos=bit.findk(rk2); } else { if(rk>me) rk--; if(rk==n-i+1) rk=1,turn++; bit.add(a[i].SE,-1); pos=bit.findk(rk); } } printf("%lld",ans); } }; int main() { srand(time(0)); mine::main(); }

来源：https://www.cnblogs.com/lijianming180/p/12286316.html