POJ 2886 Who Gets the Most Candies?(树状数组+二分)

题目链接

注意题目中给的顺序是顺时针的，所以在数组中应该是倒着存的。左就是顺时针，右就是逆时针。各种调试之后，终于A了，很多种情况考虑情况。

#include <cstring>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
char name[500001][11];
int o[500001];
int p[500001];
int s[500001];
int n;
int lowbit(int t)
{
    return t&(-t);
}
void insert(int t,int d)
{
    while(t <= n)
    {
        p[t] += d;
        t += lowbit(t);
    }
}
int getsum(int t)
{
    int sum = 0;
    while(t > 0)
    {
        sum += p[t];
        t -= lowbit(t);
    }
    return sum;
}
int find(int x)
{
    int str,mid,end,temp;
    str = 1;end = n;
    while(str < end)
    {
        mid = (str+end)/2;
        temp = getsum(mid);
        if(temp < x)
        str = mid + 1;
        else
        end = mid;
    }
    return str;
}
int main()
{
    int i,k,j,maxz,sum,sl,sr,key;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for(i = n; i >= 1; i --)
        {
            scanf("%s%d",name[i],&s[i]);
        }
        for(i = 1;i <= n;i ++)
        {
            p[i] = 0;
            o[i] = 0;
        }
        for(i = 2;i <= n;i ++)
        {
            for(j = i;j <= n;j += i)
            {
                o[j] ++;
            }
        }
        maxz = 0;
        key = 1;
        for(i = 2;i <= n;i ++)
        {
            if(maxz < o[i])
            {
                maxz = o[i];
                key = i;
            }
        }
        for(i = 1; i <= n; i ++)
        {
            insert(i,1);
        }
        k = n - k + 1;
        insert(k,-1);
        sum = n-1;
        for(i = 2;i <= key;i ++)
        {
            sl = getsum(k);
            sr = sum - sl;
            if(s[k] > 0)
            {
                s[k] = s[k]%sum;
                if(s[k] == 0) s[k] = sum;
                if(sl >= s[k])
                k = find(sl-s[k]+1);
                else
                k = find(sum-(s[k]-sl)+1);
            }
            else
            {
                s[k] = -s[k];
                s[k] = (s[k])%sum;
                if(s[k] == 0) s[k] = sum;
                if(sr >= s[k])
                k = find(sl+s[k]);
                else
                k = find(s[k]-sr);
            }
            insert(k,-1);
            sum --;
        }
        printf("%s %d
",name[k],o[key]+1);
    }
    return 0;
}