[二分答案 单调队列] BZOJ 3316 JC loves Mkk

%%%PoPoQQQ：http://blog.csdn.net/popoqqq/article/details/43059527

看到平均值最大果断二分答案

看到长度[L,R]果断单调队列

对数组维护一个前缀和，对前缀和维护单调递增的单调队列

每扫过一个数sum[i]，将sum[i-L]加入单调队列，再把距离i超过R的点删掉

长度为偶数？对奇数位置和偶数位置分别维护一个单调队列即可

每次找到大于0的子串之后记录一下分母再退出就行了

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;inline char nc(){static char buf[1000000],*p1=buf,*p2=buf;if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }return *p1++;}inline void read(int &x){char c=nc(),b=1;for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline ll GCD(ll a,ll b){return b?GCD(b,a%b):a;}int n,L,R;int maxa,a[200005];double sum[200005];ll iP,iQ,D;int Q[2][200005],l[2],r[2];inline bool Check(double mid){for (int i=1;i<=n;i++)sum[i]=sum[i-1]+a[i]-mid;l[0]=l[1]=r[0]=r[1]=-1;for (int i=L;i<=n;i++){int *Q=::Q[i&1],&r=::r[i&1],&l=::l[i&1],x=i-L;while (l<r && sum[Q[r]]>sum[x])Q[r--]=0;Q[++r]=x;while (i-Q[l+1]>R)Q[++l]=0;if(sum[i]-sum[Q[l+1]]>=0)              return iQ=i-Q[l+1];    }return 0;}inline double Bin(){double L=0,R=maxa,MID;while (R-L>=1e-6)if (Check(MID=(L+R)/2))L=MID;elseR=MID;return (L+R)/2;}int main(){freopen("t.in","r",stdin);freopen("t.out","w",stdout);read(n); read(L); read(R);if(L&1) L++; if(R&1) R--; for (int i=1;i<=n;i++)read(a[i]),a[n+i]=a[i],maxa=max(maxa,a[i]);n<<=1;iP=(ll)(Bin()*iQ+0.5);D=GCD(iQ,iP);iP/=D; iQ/=D;if (iQ==1)printf("%lld\n",iP);elseprintf("%lld/%lld\n",iP,iQ);return 0;}