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

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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，对于每个N，一开始找到小于等于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\n", name[k], fact[E]);                break;            }        }    }    return 0;}

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