题目链接：uva 11423 - Cache Simulator

题目大意；模拟一个cache，4种操作：

• ADDR x：访问一个x地址
• RANGE x y n：访问x + y * k (0≤k<n)的地址
• STAT：输出每个大小的cache，MISS的次数
• END：结束

按照从小到大的顺序给cache的大小。

解题思路：因为访问的次数不会大于1e7次，所以预先处理出访问的序列，然后对于每个位置记录前面最早出现相同值的位置。用队列记录哪些位置有存在询问。然后遍历序列，每次计算Cache需要多大才能命中，（如果小的size可以击中，那么大的size也一定可以）。
计算大小的时候除了区间长度外，还要考虑命中的次数，用树状数组维护即可。

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;
const int maxn = 1e7;
const int INF = 0x3f3f3f3f;

#define lowbit(x) ((x)&(-x))

int N, size[50], c[50];
int pos[maxn+5], fenx[maxn+5];

struct point {
    int val, mac;
    point (int val = 0, int mac = 0) {
        this->val = val;
        this->mac = mac;
    }
};

queue<int> que;
vector<point> vec;

inline bool cmp (const point& a, const point& b) {
    if (a.val != b.val)
        return a.val < b.val;
    return a.mac < b.mac;
}

int sum (int x) {
    int ret = 0;
    while (x) {
        ret += fenx[x];
        x -= lowbit(x);
    }
    return ret;
}

void add (int x, int v) {
    while (x <= maxn) {
        fenx[x] += v;
        x += lowbit(x);
    }
}

int get_distance (int i) {
    if (pos[i] == -INF)
        return INF;
    return i - pos[i] - sum(i) + sum(pos[i] - 1);
}

void hit (int i) {
    if (i <= 0)
        return;
    add(i, 1);
}

void put_ans() {
    for (int i = 1; i <= N; i++)
        printf("%d%c", c[i], i == N ? ‘\n‘ : ‘ ‘);
    memset(c, 0, sizeof(c));
}

void solve () {
    memset(c, 0, sizeof(c));
    memset(fenx, 0, sizeof(fenx));

    int n = vec.size();
    sort(vec.begin(), vec.end(), cmp);


    if (que.front() == 0) {
        put_ans();
        que.pop();
    }

    if (n == 0)
        return;
    pos[vec[0].mac] = -INF;
    for (int i = 1; i < n; i++) {
        if (vec[i].val == vec[i-1].val)
            pos[vec[i].mac] = vec[i-1].mac;
        else
            pos[vec[i].mac] = -INF;
    }

    for (int i = 1; i <= n; i++) {
        int dist = get_distance(i);
        int mv = 1;

        while (mv <= N && size[mv] < dist) {
            c[mv]++;
            mv++;
        }

        while (i == que.front()) {
            put_ans();

            que.pop();
            if (que.empty())
                break;
        }
        hit(pos[i]);
    }

    while (!que.empty())
        que.pop();
}

int main () {
    while (scanf("%d", &N) == 1) {
        for (int i = 1; i <= N; i++)
            scanf("%d", &size[i]);

        vec.clear();
        int x, y, n;
        char order[50];
        while (scanf("%s", order) == 1 && strcmp(order, "END")) {
            if (order[0] == ‘R‘) {
                scanf("%d%d%d", &x, &y, &n);
                for (int i = 0; i < n; i++)
                    vec.push_back(point(x + y * i, vec.size() + 1));
            } else if (order[0] == ‘A‘) {
                scanf("%d", &x);
                vec.push_back(point(x, vec.size() + 1));
            } else
                que.push(vec.size());
        }
        solve();
    }
    return 0;
}

uva 11423 - Cache Simulator(树状数组)