【POJ】2886 Who Gets the Most Candies?

  1 #include<cstdio>

  2 #define MAXN 500010

  3 int n,k,ans,cnt;

  4 int prime[]={2,3,5,7,11,13,17};

  5 struct node

  6 {

  7     char name[12];

  8     int val;

  9 };

 10 node p[MAXN];

 11 int tree[MAXN<<2];

 12 void AntiPrime(int res,int count,int index,int limit)

 13 {

 14     if(res<=n)

 15     {

 16         if(count>cnt)

 17         {

 18             cnt=count;

 19             ans=res;

 20         }

 21         else if(count==cnt&&res<ans)

 22             ans=res;

 23         int i,temp;

 24         for(i=temp=1;i<=limit;i++)

 25         {

 26             temp*=prime[index];

 27             if(temp*res>n)

 28                 break;

 29             else

 30                 AntiPrime(temp*res,count*(i+1),index+1,i);

 31         }

 32     }

 33 }

 34 inline void PushUp(int rt)

 35 {

 36     tree[rt]=tree[rt<<1]+tree[rt<<1|1];

 37 }

 38 void Build(int L,int R,int rt)

 39 {

 40     if(L==R)

 41         tree[rt]=1;

 42     else

 43     {

 44         int mid=(L+R)>>1;

 45         Build(L,mid,rt<<1);

 46         Build(mid+1,R,rt<<1|1);

 47         PushUp(rt);

 48     }

 49 }

 50 int Update(int x,int L,int R,int rt)

 51 {

 52     if(L==R)

 53     {

 54         tree[rt]=0;

 55         return L;

 56     }

 57     int mid=(L+R)>>1,res;

 58     if(tree[rt<<1]>=x)

 59         res=Update(x,L,mid,rt<<1);

 60     else

 61         res=Update(x-tree[rt<<1],mid+1,R,rt<<1|1);

 62     PushUp(rt);

 63     return res;

 64 }

 65 int Query(int x,int y,int L,int R,int rt)

 66 {

 67     if(x<=L&&R<=y)

 68         return tree[rt];

 69     int mid=(L+R)>>1,res=0;

 70     if(x<=mid)

 71         res+=Query(x,y,L,mid,rt<<1);

 72     if(y>mid)

 73         res+=Query(x,y,mid+1,R,rt<<1|1);

 74     return res;

 75 }

 76 int main()

 77 {

 78     int i,mov;

 79     while(~scanf("%d%d",&n,&k))

 80     {

 81         for(ans=cnt=i=1;i<=n;i++)

 82             scanf(" %s%d",p[i].name,&p[i].val);

 83         AntiPrime(1,1,0,19);

 84         Build(1,n,1);

 85         k=Update(k,1,n,1);

 86         for(i=1;i<ans;i++)

 87         {

 88             mov=Query(1,k,1,n,1);

 89             if(p[k].val>0)

 90                 mov+=p[k].val;

 91             else

 92                 mov+=p[k].val+1;

 93             mov%=tree[1];

 94             if(!mov)

 95                 mov=tree[1];

 96             else if(mov<0)

 97                 mov+=tree[1];

 98             k=Update(mov,1,n,1);

 99         }

100         printf("%s %d\n",p[k].name,cnt);

101     }

102     return 0;

103 }