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

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 #define L(rt) (rt<<1)
 8 #define R(rt) (rt<<1|1)
 9 
10 const int N=500010;
11 
12 int n,id;
13 
14 struct node{
15     int l,r,sum;    //sum 表 该区间剩余人数
16 }tree[N*3];
17 
18 struct date{
19     int val;
20     char name[15];
21 }boy[N];
22 
23 int ans[N]; //ans[i]保存第i个人跳出能得到的糖果数量
24 
25 void build(int l,int r,int rt){
26     tree[rt].l=l;
27     tree[rt].r=r;
28     tree[rt].sum=r-l+1;
29     if(l==r)
30         return ;
31     int mid=(l+r)>>1;
32     build(l,mid,L(rt));
33     build(mid+1,r,R(rt));
34 }
35 
36 int update(int key,int rt){
37     tree[rt].sum--;
38     if(tree[rt].l==tree[rt].r)
39         return tree[rt].l;
40     if(tree[L(rt)].sum>=key)
41         return update(key,L(rt));
42     else
43         return update(key-tree[L(rt)].sum,R(rt));
44 }
45 
46 void Solve(){   //计算ans
47     memset(ans,0,sizeof(ans));
48     for(int i=1;i<=n;i++){
49         ans[i]++;
50         for(int j=2*i;j<=n;j+=i)
51             ans[j]++;
52     }
53     int max=ans[1];
54     id=1;
55     for(int i=2;i<=n;i++)   //找出第几个人跳出获得的糖最多
56         if(ans[i]>max){
57             max=ans[i];
58             id=i;
59         }
60 }
61 
62 int main(){
63 
64     //freopen("input.txt","r",stdin);
65 
66     int i,k,mod;
67     while(~scanf("%d%d",&n,&k)){
68         Solve();
69         for(i=1;i<=n;i++)
70             scanf("%s%d",boy[i].name,&boy[i].val);
71         build(1,n,1);
72         mod=tree[1].sum;
73         int pos=0;
74         boy[0].val=0;
75         n=id;
76         while(n--){
77             if(boy[pos].val>0)  //k表剩余的人中从左起第k中出队
78                 k=((k-1+boy[pos].val-1)%mod+mod)%mod+1;
79             else
80                 k=((k-1+boy[pos].val)%mod+mod)%mod+1;
81             pos=update(k,1);
82             mod=tree[1].sum;
83         }
84         printf("%s %d\n",boy[pos].name,ans[id]);
85     }
86     return 0;
87 }