hdu 3486 Interviewe 二分+RMQ

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>using namespace std;const int maxn=200010;int dp[20][maxn];//dp[i][j]表示从第j个元素开始，连续的2^i个元素中的最大值int a[maxn],n,m;int lg[maxn];//lg[i]求表示2^t=i,lg[i]=floor(t);void init(){    int i,j,t;    lg[0]=-1;    for(i=1;i<maxn;i++)        lg[i]=lg[i/2]+1;    for(i=1;i<=n;i++)        dp[0][i]=a[i];    for(i=1;i<=lg[n];i++)    {        t=n+1-(1<<i);//注意dp[i][j]求得是[j,j+(1<<i)-1],所以要+1        for(j=1;j<=t;j++)            dp[i][j]=max(dp[i-1][j],dp[i-1][j+(1<<(i-1))]);    }}int rmq(int l,int r){    int t=lg[r-l+1];    return max(dp[t][l],dp[t][r-(1<<t)+1]);}int find(int x,int y){    int i,ans=0;    for(i=1;i<=y;i++)    {        ans+=rmq((i-1)*x+1,i*x);        if(ans>m)return ans;    }    return ans;}/*int get_in(){    int ans=0;    char ch;    while((ch=getchar())==' '||c=='\n');    ans=ch-'0';    while((c=getchar())<='9'&&c>='0')        ans=ans*10+ch-'0';    return ans;}*/int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        if(n<0&&m<0)break;        int i,j,k,p=0,q=0;        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            if(a[i]>p)p=a[i];            q+=a[i];        }        if(p>m){printf("1\n");continue;}        if(q<=m){printf("-1\n");continue;}        init();        int l=1,r=n,mid,t;        int ans=0;        while(l<=r)        {            mid=(l+r)/2;            t=find(n/mid,mid);            if(t>m)            {                r=mid-1;                ans=mid;            }            else                l=mid+1;        }        printf("%d\n",ans);    }    return 0;}/*    二分+RMQ，    RMQ(l,r)，求区间[l,r]中的最大(小)值。*/