挑战练习题3.3 POJ 2886 Who Gets the Most Candies? 树状数组
2017-03-24 19:32
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题目链接:
http://poj.org/problem?id=2886题意:
抢糖:N个熊孩子围成一个圈,从第K个开始淘汰,每淘汰一个,出示手中的数字,决定下一个淘汰者,正数表示左手第n个,负数反之。每个人可以拿到的存活回数的因数个数的糖果,求拿到最多糖果数的孩子的名字以及糖果数。题解:
http://www.hankcs.com/program/algorithm/poj-2886-who-gets-the-most-candies.html代码:
#include <iostream> #include <cstring> #include <cstdio> using namespace std; typedef long long ll; #define MS(a) memset(a,0,sizeof(a)) #define MP make_pair #define PB push_back const int INF = 0x3f3f3f3f; const ll INFLL = 0x3f3f3f3f3f3f3f3fLL; inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } ////////////////////////////////////////////////////////////////////////// const int maxn = 5e5+10; int factor_table[maxn]; char name[maxn][16]; int integer[maxn]; void init_factor_table(int n){ fill(factor_table,factor_table+n+1,1); for(int i=2; i<=n; i++){ if(factor_table[i]==1){ for(int j=i; j<=n; j+=i){ int k = 0; for(int m=j; m%i==0; m/=i,++k); factor_table[j] *= (k+1); } } } } struct BinaryIndexTree{ int bit[maxn]; int size; void init(int size){ this->size = size; MS(bit); } int sum(int i){ int s = 0; while(i > 0){ s += bit[i]; i -= i&-i; } return s; } void add(int i,int v){ ++i; while(i < maxn){ bit[i] += v; i += i&-i; } } int binary_search(int id){ int l=0,r=size,res=0; while(l<=r){ int mid = (l+r)/2; if(sum(mid) <= id) res=mid,l=mid+1; // mid是最初的编号 else r=mid-1; } return res; } }T; int main(){ int N,K; init_factor_table(maxn); while(cin>>N>>K){ --K; T.init(N); for(int i=0; i<N; i++){ scanf("%s %d",name[i],&integer[i]); T.add(i,1); } int candies=-1,index; for(int i=0; i<N; i++){ if(candies < factor_table[i+1]){ candies = factor_table[i+1]; index = K; } T.add(K,-1); if(i < N-1){ int mod = N-i-1; int id = T.sum(K)+integer[K]+(integer[K]>0 ? -1 : 0); // T.sum(K)表示第K个编号的人是当前轮的第几个。id是下一个剔除的人排在当前序列中的第几个 id = (id%mod+mod)%mod; // id可能是负的很大的数 K = T.binary_search(id); // 找到下一个剔除的最初编号 } } cout << name[index] << " " << candies << endl; } return 0; }
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