线段树(单点更新) POJ 2886 Who Gets the Most Candies?

 1 #include <cstdio>
 2 #include <cstring>
 3 #define lson l, m, rt << 1
 4 #define rson m+1, r, rt << 1 | 1
 5 
 6 const int MAX_N = 500000 + 10;
 7 int sum[MAX_N << 2];
 8 struct node
 9 {
10     char name[15];
11     int val;
12 }boy[MAX_N<<2];
13 int ans[MAX_N];
14 int id;
15 int n, k;
16 
17 void build(int l, int r, int rt)
18 {
19     sum[rt] = r - l + 1;
20     if (l == r)
21         return ;
22     int m = (l + r) >> 1;
23     build (lson);
24     build (rson);
25 }
26 
27 int update(int p, int l, int r, int rt)
28 {
29     sum[rt]--;
30     if (l == r)
31     {
32         return l;
33     }
34     int m = (l + r) >> 1;
35     if (p <= sum[rt<<1])
36     {
37         return update (p, lson);
38     }
39     else
40     {    
41         return update (p - sum[rt<<1], rson);
42     }
43 }
44 
45 void Solve(){   //计算ans
46     memset(ans,0,sizeof(ans));
47     for(int i=1;i<=n;i++){
48         ans[i]++;
49         for(int j=2*i;j<=n;j+=i)
50             ans[j]++;
51     }
52     int max=ans[1];
53     id=1;
54     for(int i=2;i<=n;i++)   //找出第几个人跳出获得的糖最多
55         if(ans[i]>max){
56             max=ans[i];
57             id=i;
58         }
59 }
60 
61 int main(void)      //POJ 2886 Who Gets the Most Candies?
62 {
63     //freopen ("inF.txt", "r", stdin);
64 
65     while (~scanf ("%d%d", &n, &k))
66     {
67         build (1, n, 1);
68         for (int i=1; i<=n; ++i)
69         {
70             scanf ("%s%d", &boy[i].name, &boy[i].val);
71             //printf ("%s%d\n", boy[i].name, boy[i].val);
72         }
73         Solve();
74         //int maxn = f(n);
75         int mod = sum[1];
76         boy[0].val = 0;
77         int pos = 0;
78         //printf ("%d\n", id);
79         int m=id;
80         while (m--)
81         {
82             if (boy[pos].val > 0)
83                 k=((k-1+boy[pos].val-1)%mod+mod)%mod+1;
84             else
85                 k=((k-1+boy[pos].val)%mod+mod)%mod+1;
86             pos = update (k, 1, n, 1);
87             //printf ("%d ", pos);
88             mod = sum[1];
89         }
90         printf ("%s %d\n", boy[pos].name, ans[id]);
91     }
92 }