poj 2886 Who Gets the Most Candies?(线段树)

题目链接：poj 2886 Who Gets the Most Candies?

题目大意：N个人围成一圈玩约瑟夫环游戏，不同的是。步长不固定，由前一个出局的人决定。给定K表示起始的人。

第i个淘汰的人将获得g(i)个糖果，问说谁获得的糖果最多。

g(x)为x的因子个数。

解题思路：起始g(x)是成阶段的，所以打表处理处g(x)递增值，对于每一个N，一開始找到小于等于N的最大x，那么第x个淘汰的人即为获得糖果数最多的家伙。剩下的就用线段树模拟游戏过程。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 500005;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)+1)

struct Node {
    int l, r, s;
    void set (int l, int r, int s) {
        this->l = l;
        this->r = r;
        this->s = s;
    }
}nd[maxn * 4];

const int antipri[36] = {1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,20160,25200,27720,45360,50400,55440,83160,110880,166320,221760,277200,332640,498960,500001};
const int fact[36] = {1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60,64,72,80,84,90,96,100,108,120,128,144,160,168,180,192,200};

int N, B, v[maxn];
char name[maxn][20];

void build (int u, int l, int r) {
    nd[u].set(l, r, r - l + 1);

    if (l == r) 
        return ;

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
}

int query(int u, int x) {
    nd[u].s--;

    if (nd[u].l == nd[u].r)
        return nd[u].l;

    if (nd[lson(u)].s >= x)
        return query(lson(u), x);
    else
        return query(rson(u), x - nd[lson(u)].s);
}

int bsearch (int x) {
    int l = 0, r = 35;
    for (int i = 0; i < 20; i++) {
        int mid = (l + r) / 2;
        if (antipri[mid] > x)
            r = mid;
        else
            l = mid;
    }
    return l;
}

int main () {
    while (scanf("%d%d", &N, &B) == 2) {
        for (int i = 1; i <= N; i++)
            scanf("%s%d", name[i], &v[i]);
        build(1, 1, N);

        v[0] = 0;
        int E = bsearch(N), k = 0;
        for (int i = N; i; i--) {
            B = ((B + v[k] - (v[k] > 0 ? 2 : 1)) % i + i) % i + 1;
            k = query(1, B);

            if (N - i + 1 == antipri[E]) {
                printf("%s %d
", name[k], fact[E]);
                break;
            }
        }
    }
    return 0;
}