2793 [Poi2012]Vouchers

我们直接模拟就可以了= =

now[x]表示x的倍数已经取到x * i了，于是每次读入x，直接向上枚举x个没取过的数即可。

 /**************************************************************
     Problem: 2793
     User: rausen
     Language: C++
     Result: Accepted
     Time:5296 ms
     Memory:14476 kb
 ****************************************************************/
 
 #include <cstdio>
 #include <algorithm>
 
 using namespace std;
 typedef long long ll;
 const int N = ;
 
 int n, m, mx;
 int now[N];
 ll tot, ans[N], cnt;
 bool f[N], vis[N];
 
 inline int read() {
     int x = , sgn = ;
     char ch = getchar();
     while (ch < '' || '' < ch) {
         if (ch == '-') sgn = -;
         ch = getchar();
     }
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return sgn * x;
 }
 
 int main() {
     int i, x, t, C;
     for (m = read(), i = ; i <= m; ++i) {
         x = read();
         f[x] = , mx = max(mx, x);
     }
 
     n = read();
     while (n--) {
         x = t = read();
         for (i = now[x] + x, C = ; i <= mx; i += x) {
             if (!vis[i]) {
                 ++cnt, --t, vis[i] = ;
                 if (f[i]) ans[++tot] = cnt;
                 if (++C == x) break;
             }
             now[x] = i;
         }
         cnt += t;
     }
     printf("%lld\n", tot);
     for (i = ; i <= tot; ++i)
         printf("%lld\n", ans[i]);
 }

（p.s. 没有的搞懂，一开始各种WA，把ans数组改成long long就AC了？蒟蒻跪求大神求教）