POJ (线段树) Who Gets the Most Candies?

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cmath>
 4 #include <string>
 5 #include <algorithm>
 6 #include <vector>
 7 using namespace std;
 8 
 9 const int maxn = 500000 + 10;
10 int n, k;
11 int sum[maxn << 2];
12 
13 void build(int o, int L, int R)
14 {
15     if(L == R) { sum[o] = 1; return; }
16     int M = (L + R) / 2;
17     build(o*2, L, M);
18     build(o*2+1, M+1, R);
19     sum[o] = sum[o*2] + sum[o*2+1];
20 }
21 
22 int update(int o, int L, int R, int p)
23 {
24     if(L == R) { sum[o] = 0; return L; }
25     int M = (L + R) / 2;
26     int ans;
27     if(sum[o*2] >= p) ans = update(o*2, L, M, p);
28     else ans = update(o*2+1, M+1, R, p-sum[o*2]);
29     sum[o] = sum[o*2] + sum[o*2+1];
30     return ans;
31 }
32 
33 const int maxp = 750;
34 int prime[maxp], pcnt = 0;
35 bool vis[maxp];
36 
37 void prime_table()
38 {
39     int m = sqrt(maxp);
40     for(int i = 2; i <= m; i++) if(!vis[i])
41         for(int j = i * i; j < maxp; j += i) vis[j] = true;
42     for(int i = 2; i < maxp; i++) if(!vis[i]) prime[pcnt++] = i;
43 }
44 
45 int F(int n)
46 {
47     int ans = 1;
48     for(int i = 0; i < pcnt && n > 1; i++) if(n % prime[i] == 0)
49     {
50         int c = 0;
51         while(n % prime[i] == 0) { n /= prime[i]; c++; }
52         ans *= c + 1;
53     }
54     if(n > 1) ans <<= 1;
55     return ans;
56 }
57 
58 int next[maxn];
59 char stu[maxn][11];
60 
61 int main()
62 {
63     //freopen("in.txt", "r", stdin);
64 
65     prime_table();
66 
67     while(scanf("%d%d", &n, &k) == 2)
68     {
69         build(1, 0, n - 1);
70         int _max = 0, id;
71         for(int i = 0; i < n; i++) { scanf("%s%d", stu[i], &next[i]); }
72 
73         int pos = k - 1;
74         for(int i = 1; i <= n; i++)
75         {
76             int t = update(1, 0, n - 1, pos + 1);
77             int f = F(i);
78             if(f > _max)
79             {
80                 _max = f;
81                 id = t;
82             }
83             if(i == n) break;
84             int MOD = n - i;
85             if(next[t] < 0) next[t]++;  //负数的情况还要有一点小改动，在这坑了好久
86             pos = (((pos + next[t] - 1) % MOD) + MOD) % MOD;
87         }
88         printf("%s %d\n", stu[id], _max);
89     }
90 
91     return 0;
92 }