poj2886 Who Gets the Most Candies?

思路：

先打反素数表，即可确定因子最多的那个数。然后模拟踢人的过程确定对应的人名。模拟的过程使用线段树优化加速。

实现：

 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 using namespace std;

 const int MAXN = ;

 int antiprime[] = {, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , };
 int fac[] = {, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , };

 int tree[MAXN << ], num[MAXN], n, k;
 char name[MAXN][];

 void build(int num, int l, int r)
 {
     if (l == r) { tree[num] = ; return; }
     int m = l + r >> ;
     build(num << , l, m);
     build(num <<  | , m + , r);
     tree[num] = tree[num << ] + tree[num <<  | ];
 }

 int kick(int num, int l, int r, int x) // 在[l, r]范围内踢掉第x个人
 {
     if (l == r) { tree[num] = ; return l; }
     int m = l + r >> , tmp = tree[num << ], ans = -;
     if (tmp >= x) ans = kick(num << , l, m, x);
     else ans = kick(num <<  | , m + , r, x - tmp);
     tree[num] = tree[num << ] + tree[num <<  | ];
     return ans;
 }

 int main()
 {
     while (scanf("%d %d", &n, &k) != EOF)
     {
         for (int i = ; i <= n; i++) scanf("%s %d", name + i, num + i);
         build(, , n);
         int now = k, out = -, cnt = n;
         int tmp = lower_bound(antiprime, antiprime + , n) - antiprime;
         int w = antiprime[tmp] > n ? tmp -  : tmp;
         for (int i = ; i < antiprime[w]; i++)
         {
             out = kick(, , n, now);
             cnt--;
             if (!cnt) break;
             if (num[out] > ) now = ((now + num[out] - ) % cnt + cnt) % cnt;
             else now = ((now + num[out]) % cnt + cnt) % cnt;
             now = !now ? cnt : now;
         }
         printf("%s %d\n", name[out], fac[w]);
     }
     return ;
 }