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 = 1000005;
 
int n, m, mx;
int now[N];
ll tot, ans[N], cnt;
bool f[N], vis[N];
 
inline int read() {
    int x = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || '9' < ch) {
        if (ch == '-') sgn = -1;
        ch = getchar();
    }
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return sgn * x;
}
 
int main() {
    int i, x, t, C;
    for (m = read(), i = 1; i <= m; ++i) {
        x = read();
        f[x] = 1, mx = max(mx, x);
    }
     
    n = read();
    while (n--) {
        x = t = read();
        for (i = now[x] + x, C = 0; i <= mx; i += x) {
            if (!vis[i]) {
                ++cnt, --t, vis[i] = 1;
                if (f[i]) ans[++tot] = cnt;
                if (++C == x) break;
            }
            now[x] = i;
        }
        cnt += t;
    }
    printf("%lld
", tot);
    for (i = 1; i <= tot; ++i)
        printf("%lld
", 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）