POJ - 2886 Who Gets the Most Candies

筛法+划分树。

枚举因子，类似筛法计算因子数量，复杂度为n/2 + n/3 + ... + n/n ≈ O(nlogn)。

值域已知，只有删除操作，寻找kth，用划分树就好了O(nlogn)。二分+树状数组也可以，只是复杂度多乘一个logn。

/*********************************************************
*            ------------------                          *
*   author AbyssalFish                                   *
**********************************************************/
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<numeric>
using namespace std;

const int MAX_N = 5e5+1, LEN = 11;

char name[MAX_N][LEN];
int card[MAX_N];

int N, K;

int C[MAX_N];

int q_Kth(int k)
{
int p = 0, c = 0;
for(int i = 19; i >= 0; i--){
p += 1<<i;
(p >= N || c + C[p] >= k)? p -= 1<<i : c += C[p];
}
return p+1;
}

void sub(int x)
{
while(x <= N){
C[x]--; x += x&-x;
}
}

int Fp[MAX_N+1];

//#define LOCAL
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
//cout<<(1<<20);//ceil(log2(5e5)+1);
scanf("%d%d",&N,&K);
for(int i = 2; i <= N; i++){
for(int j = i; j <= N; j += i){
Fp[j]++;
}
}
int tar = 1;
for(int i = 2; i <= N; i++){
if(Fp[tar] < Fp[i]) tar = i;
}
for(int i = 0; i < N; i++) scanf("%s%d", name[i], card+i);
for(int i = 1; i <= N; i++) {
C[i] = i&-i;
}
int pos , k = K-1, mn = N-tar;
for(int n = N; n-- > mn; ){
pos = q_Kth(k+1)-1; sub(pos+1);
if(n){
if(card[pos] > 0) card[pos]--;
k = (card[pos]+k)%n;
if(k < 0) k += n;
}
}
printf("%s %d\n", name[pos], Fp[tar]+1);
return 0;
}

树状数组