POJ-2886 Who Gets the Most Candies?---线段树+约瑟夫环

 1 #include<iostream>
 2 #include<cstdio>
 3 #define MID(l, r) (l + (r - l) / 2)
 4 #define lson(o) (o * 2)
 5 #define rson(o) (o * 2 + 1)
 6 using namespace std;
 7 typedef long long ll;
 8 const int INF = 1e9 +7;
 9 const int maxn = 2e6 + 10;
10 int  h, w, n;
11 struct node
12 {
13     int l, r, sum;
14 }tree[maxn];
15 int ans[maxn];
16 void build(int o, int l, int r)
17 {
18     tree[o].l = l, tree[o].r = r;
19     if(l == r)
20     {
21         tree[o].sum = 1;
22         return;
23     }
24     int m = MID(l, r);
25     int lc = lson(o), rc = rson(o);
26     build(lc, l, m);
27     build(rc, m + 1, r);
28     tree[o].sum = tree[lc].sum + tree[rc].sum;
29 }
30 int id;
31 void query(int o, int a)//查询第a个数下标，同时删除这个数
32 {
33     if(tree[o].l == tree[o].r)
34     {
35         id = tree[o].l;
36         tree[o].sum = 0;
37         return;
38     }
39     int lc = lson(o), rc = rson(o);
40     if(a <= tree[lc].sum)query(lc, a);
41     else query(rc, a - tree[lc].sum);
42     tree[o].sum = tree[lc].sum + tree[rc].sum;
43 }
44 int divisor_num[maxn];
45 void init(int n)
46 {
47     for(int i = 1; i <= n; i++)
48     {
49         divisor_num[i]++;
50         for(int j = i * 2; j <= n; j += i)
51             divisor_num[j]++;
52     }
53 }
54 int solve(int n)
55 {
56     int max = 0, maxid;
57     for(int i = 1; i <= n; i++)
58         if(divisor_num[i] > max)
59     {
60         max = divisor_num[i];
61         maxid = i;
62     }
63     return maxid;
64 }
65 char name[500005][11];
66 int a[maxn];
67 int main()
68 {
69     int n, m;
70     init(500000);
71     while(scanf("%d%d", &n, &m) != EOF)
72     {
73         for(int i = 1; i <= n; i++)
74             scanf("%s%d", name[i], &a[i]);
75         build(1, 1, n);
76         int ansid = solve(n), mod = n;
77         int t = ansid;
78         while(1)
79         {
80             query(1, m);
81             if(--t == 0)break;
82             mod--;
83             if(a[id] > 0)
84             {
85                 m = m - 1 + a[id];
86                 m = ((m - 1) % mod + mod) % mod + 1;
87             }
88             else
89             {
90                 m = m + a[id];
91                 m = ((m - 1) % mod + mod) % mod + 1;
92             }
93         }
94         printf("%s %d\n", name[id], divisor_num[ansid]);
95     }
96     return 0;
97 }