POJ 2886 Who Gets the Most Candies?
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 8212 | Accepted: 2475 | |
| Case Time Limit: 2000MS | ||
Description
N children are sitting in a circle to play a game.
The children are numbered from 1 to N in clockwise order. Each of them has a card with a non-zero integer on it in his/her hand. The game starts from the K-th child, who tells all the others the integer on his card and jumps out of the circle. The integer on his card tells the next child to jump out. Let A denote the integer. If A is positive, the next child will be the A-th child to the left. If A is negative, the next child will be the (−A)-th child to the right.
The game lasts until all children have jumped out of the circle. During the game, the p-th child jumping out will get F(p) candies where F(p) is the number of positive integers that perfectly divide p. Who gets the most candies?
Input
Output
Output one line for each test case containing the name of the luckiest child and the number of candies he/she gets. If ties occur, always choose the child who jumps out of the circle first.
Sample Input
4 2 Tom 2 Jack 4 Mary -1 Sam 1
Sample Output
Sam 3
Source
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#define N 500010
#define M 15
using namespace std;
bool ch[N];
int a[N],b[N],sum[N];
struct num
{
char s1[M];
int val;
}in[N];
struct tree
{
int l,r,sum;
}d[N*4];
int n,m,res,ans;
int main()
{
//freopen("data.in","r",stdin);
void pre_build(int k,int l,int r);
void get(int k,int l,int r);
void deal(int k,int tar);
memset(ch,true,sizeof(ch));
for(int i=2;i<=N-10;i++)
{
if(ch[i])
{
for(int j=2;i*j<=N-10;j++)
{
ch[i*j]=false;
}
}
}
int Top=0;
for(int i=2;i<=N-10;i++)
{
if(ch[i])
{
a[Top++]=i;
}
}
sum[1]=1;
memset(b,0,sizeof(b));
for(int i=2;i<=N-10;i++)
{
int co=i,x;
for(int j=0;j<=Top-1;j++)
{
if(ch[co])
{
b[j]++;
x=j;
break;
}
while(co%a[j]==0)
{
b[j]++;
x=j;
co=co/a[j];
}
if(co==1)
{
break;
}
}
int s =1;
for(int j=0;j<=x;j++)
{
s=s*(b[j]+1);
b[j]=0;
}
sum[i]=s;
}
while(scanf("%d %d",&n,&m)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%s %d",in[i].s1,&in[i].val);
}
int Max,pt;
pre_build(1,1,n);
Max=sum[1];
pt=m;
for(int i=1;i<=n;i++)
{
if(i==1)
{
int s=m%n;
if(s==0)
{
s=n;
}
deal(1,s);
if(Max<sum[i])
{
Max=sum[i];
pt=res;
}
continue;
}
ans=0;
get(1,1,res);
int uv;
if(in[res].val<0)
{
uv = (in[res].val*-1)%(n-i+1);
if(uv==0)
{
uv=n-i+1;
}
uv = (n-i+1)-uv+1+ans;
uv = uv%(n-i+1);
}else
{
uv = (in[res].val)%(n-i+1);
uv = uv+ans;
uv = uv%(n-i+1);
}
if(uv==0)
{
uv = n-i+1;
}
deal(1,uv);
if(Max<sum[i])
{
Max=sum[i];
pt=res;
}
}
printf("%s %d\n",in[pt].s1,Max);
}
return 0;
}
void pushup(int k)
{
d[k].sum = d[k*2].sum+d[k*2+1].sum;
}
void pre_build(int k,int l,int r)
{
d[k].l = l;
d[k].r = r;
if(l==r)
{
d[k].sum=1;
return ;
}
int mid=(l+r)/2;
pre_build(k*2,l,mid);
pre_build(k*2+1,mid+1,r);
pushup(k);
}
void get(int k,int l,int r)
{
if(d[k].l==l&&d[k].r==r)
{
ans+=d[k].sum;
return ;
}
int mid=(d[k].l+d[k].r)/2;
if(mid>=r)
{
get(k*2,l,r);
}else if(mid<l)
{
get(k*2+1,l,r);
}else
{
get(k*2,l,mid);
get(k*2+1,mid+1,r);
}
}
void deal(int k,int tar)
{
if(d[k].l==d[k].r)
{
res=d[k].l;
d[k].sum=0;
return ;
}
if(d[k*2].sum<tar)
{
deal(k*2+1,tar-d[k*2].sum);
}else
{
deal(k*2,tar);
}
pushup(k);
}