URAL 2080 Wallet

找规律发现只要找到两个相同数字之间，有多少个不同的数字，即为答案。

可以用树状数组离线处理。

坑点是卡有很多张，没用完的情况，后面的卡直接放在哪里，

就是

10 5

1 2 3 4 5 这样

开始数据要输出到10

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>

const int maxn = 1e5 + 20;
int n,m;
int c[maxn];//树状数组，多case的记得要清空
int lowbit (int x)//得到x二进制末尾0的个数的2次方 2^num
{
    return x&(-x);
}
void add (int pos,int val)//在第pos位加上val这个值
{
    while (pos<=m) { //n是元素的个数
        c[pos] += val;
        pos += lowbit(pos);
    }
    return ;
}
int get_sum (int pos) //求解：1--pos的总和
{
    int ans = 0;
    while (pos) {
        ans += c[pos];
        pos -= lowbit(pos);
    }
    return ans;
}
int ans[maxn];
int a[maxn];
struct node {
    int L,R,id;
    bool operator < (const node &rhs) const {
        return R < rhs.R;
    }
}query[maxn];
int ansQ[maxn];
int book[maxn];
int cur[maxn];
vector<int>pos[maxn];
void work ()
{
    int lenans = 0;
    scanf ("%d%d", &n, &m);
    for (int i = 1; i <= m; ++i) scanf ("%d", &a[i]);
    for (int i = 1; i <= n; ++i) cur[i] = 1;
    for (int i = 1; i <= m; ++i) {
        if (book[a[i]] == 0) {
            ans[++lenans] = a[i];
            book[a[i]] = 1;
        }
        pos[a[i]].push_back(i);
    }
    for (int i = 1; i <= n; ++i) {
        if (book[i] == 0) {
            ans[++lenans] = i;
        }
    }
    memset (book, 0, sizeof book);
    for (int i = 1; i <= m; ++i) {
        if (pos[a[i]].size() <= cur[a[i]]) {
            query[i].L = i; query[i].R = m; query[i].id = i;
        } else {
            query[i].L = i; query[i].R = pos[a[i]][cur[a[i]]];
            query[i].id = i;
        }
        cur[a[i]]++;
    }
    sort (query + 1, query + 1 + m);
    int now = 1;
    for (int i = 1; i <= m; ++i) {
        for (int j = now; j <= query[i].R; ++j) {
            if (book[a[j]]) {
                add(book[a[j]], -1);
            }
            book[a[j]] = j;
            add(j,1);
        }
        now = query[i].R + 1;
        ansQ[query[i].id] = get_sum(query[i].R) - get_sum(query[i].L - 1) - 1;
    }
    for (int i = 1; i <= lenans; ++i) {
        printf ("%d ",ans[i]);
    }
    printf ("\n");
    for (int i = 1; i <= m; ++i) {
        printf ("%d\n",ansQ[i]);
    }
    return ;
}
int main()
{
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    work ();
    return 0;
}