lightoj 1097

Lucky numbers are defined by a variation of the well-known sieve of Eratosthenes. Beginning with the natural numbers strike out all even ones, leaving the odd numbers 1, 3, 5, 7, 9, 11, 13, ...The second number is 3, next strike out every third number, leaving 1, 3, 7, 9, 13, ... The third number is 7, next strike out every seventh number and continue this process infinite number of times. The numbers surviving are called lucky numbers. The first few lucky numbers are:

1, 3, 7, 9, 13, 15, 21, 25, 31, 33, ...

In this problem your task is to find the n^th lucky number where n is given in input.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 10⁵).

Output

For each case, print the case number and the n^th lucky number.

题意：先给出一个由奇数组成的数列, 1 3 5 7 9 ….. ,现在告诉你, 第i次操作从序列中取出第i个数, 然后每隔a[i]去掉一个数, 问你最后幸存下来的第n个数是多少。

利用线段数暴力操作一遍就行了，由于样例很友善都给出了第100000个数是1429431所以建一个大小为1429431的树就可以了，但是节点大小不要设为

(1429431) << 2这样会ME的。1429431小与2^21所以总结点数小于2^22，大概是1429431的3倍左右就可以了。

至于怎么操作数，要用到前缀和的思想，区间的和表示这区间总公能放多少的点，删掉一个就想这个点的值改为0。具体递归为

if T[p].num < pos  (pos - T[p].num , (p<<1)|1) else (pos , p<<1)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int M = 1429431;
struct TnT {
    int l , r , sum;
}T[M * 3];
void build(int l , int r , int p) {
    int mid = (l + r) >> 1;
    T[p].l = l , T[p].r = r;
    if(T[p].l == T[p].r) {
        T[p].sum = 1;
        if(T[p].l % 2 == 0)
            T[p].sum = 0;
        return ;
    }
    build(l , mid , p << 1);
    build(mid + 1 , r , (p << 1) | 1);
    T[p].sum = T[p << 1].sum + T[(p << 1) | 1].sum;
}
void updata(int pos , int p) {
    if(T[p].l == T[p].r) {
        T[p].sum = 0;
        return ;
    }
    if(T[p << 1].sum < pos) {
        updata(pos - T[p << 1].sum , (p << 1) | 1);
    }
    else {
        updata(pos , p << 1);
    }
    T[p].sum = T[p << 1].sum + T[(p << 1) | 1].sum;
}
int query(int pos , int p) {
    if(T[p].l == T[p].r) {
        return T[p].l;
    }
    if(T[p << 1].sum < pos) {
        return query(pos - T[p << 1].sum , (p << 1) | 1);
    }
    else {
        return query(pos , p << 1);
    }
}
void init() {
    build(1 , M , 1);
    int gg;
    for(int i = 2 ; i <= T[1].sum ; i++) {
        gg = query(i , 1);
        for(int j = gg ; j <= T[1].sum ; j += (gg - 1)) {
            updata(j , 1);
        }
    }
}
int main()
{
    int t;
    init();
    scanf("%d" , &t);
    int ans = 0;
    while(t--) {
        ans++;
        int n;
        scanf("%d" , &n);
        printf("Case %d: %d
" , ans , query(n , 1));
    }
    return 0;
}