题意:变形的约瑟夫环模型,每个人有一个数字a,从第K个人开始出列,如果数字是正的,就往后数a个人出列,如果书负数,就往反方向数。
然后用最基本的线段树处理约瑟夫环的方法即可
但是题目要求的是第x个出列的人的名字,x为1-N中约数最多的数中的最小的那个。这里需要求反素数,即不大于N约数最多的。
写起来比较多,容易写错,一开始连素数打表都写作了QAQ
#include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <cstdlib> #include <map> #include <set> #include <vector> #include <string> #include <queue> #include <stack> #include <climits> using namespace std; typedef long long LL; const int maxn = 500000 + 5; int sum[maxn << 2]; //线段树 #define lson rt<<1,l,mid #define rson rt<<1|1,mid+1,r void build(int rt,int l,int r) { if(l == r) sum[rt] = 1; else { int mid = (l + r) >> 1; build(lson); build(rson); sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } } void update(int rt,int l,int r,int tar,int x) { if(l == r) sum[rt] = x; else { int mid = (l + r) >> 1; if(tar <= mid) update(lson,tar,x); else update(rson,tar,x); sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } } int query(int rt,int l,int r,int v) { if(l == r) return l; else { int mid = (l + r) >> 1,lc = rt<<1,rc = lc + 1; if(sum[lc] >= v) return query(lson,v); else return query(rson,v - sum[lc]); } } //反素数生成 int prime[20] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71}; int minv[maxn],times[40]; void dfs(LL arr,int nowcnt,int nowt,LL lim) { if(arr > lim) return; //cout << arr << " " << nowcnt << " " << nowt << " " << lim << endl; minv[nowcnt] = min(minv[nowcnt],(int)arr); for(int i = 1;nowt == 0 || i <= times[nowt - 1];i++) { arr *= prime[nowt]; if(arr > lim) break; nowcnt = nowcnt / i * (i + 1); times[nowt] = i; dfs(arr,nowcnt,nowt + 1,lim); } } char name[maxn][20]; int k[maxn],N,K; void solve() { int nowpos = K,ans,ansv,nowk,pos = K; build(1,1,N); for(int i = 1;i <= N;i++) if(minv[i] <= N) ansv = i; for(int i = 1;i <= N;i++) { pos = query(1,1,N,nowpos); update(1,1,N,pos,0); // printf("%d:%s out in %d\n",pos,name[pos],i); nowk = k[pos]; if(i == minv[ansv]) { ans = pos; break; } if(nowk > 0) nowpos = ((nowpos + nowk - 2) % (N - i) + (N - i)) % (N - i) + 1; else nowpos = ((nowpos + nowk - 1) % (N - i) + (N - i)) % (N - i) + 1; } printf("%s %d\n",name[ans],ansv); } void setfile() { freopen("in.txt","r",stdin); freopen("a.txt","w",stdout); } int main() { // setfile(); for(int i = 1;i <= maxn;i++) minv[i] = INT_MAX; dfs(1,1,0,maxn); while(scanf("%d%d",&N,&K) != EOF) { for(int i = 1;i <= N;i++) scanf("%s%d",name[i],&k[i]); solve(); } return 0; }
POJ 2886 Who Gets the Most Candies? 反素数+线段树,布布扣,bubuko.com
POJ 2886 Who Gets the Most Candies? 反素数+线段树
原文地址:http://www.cnblogs.com/rolight/p/3841634.html