CodeForces 518C - Watto and Mechanism（模拟）

题意：有n（1 <= n <= 10^5）个应用，每屏有k（1 <= k <= 10^5）个应用，现在有m（1 <= m <= 10^5）个操作，每次操作会使用一个应用（使用时需滑到应用所在的屏），使用后此应用与前边的相邻应用交换位置，退出此应用后会回到初始屏。问这m次操作总的滚动屏幕次数。

模拟即可。

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
typedef long long ll;
typedef unsigned long long llu;
const int MAXN = 100 + 10;
const int MAXT = 100000 + 10;
const int INF = 0x7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
using namespace std;

int n, m, k, a[MAXT], loc[MAXT][2];
vector<int> g[MAXT];



int main(){
    memset(loc, -1, sizeof loc);
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 0; i < n; ++i)  scanf("%d", a + i);
    int lur = 0;
    for(int i = 0; i < n; ++i){
        int j;
        for(j = 0; j < k && i + j < n; ++j){
            g[lur].push_back(a[i + j]);
            loc[a[i + j]][0] = lur;
            loc[a[i + j]][1] = j;
        }
        ++lur;
        i = i + j - 1;
    }
    llu ans = 0;
    while(m--){
        scanf("%d", &lur);
        ans += (llu)(loc[lur][0] + 1);
        if(loc[lur][1] == 0){
            if(loc[lur][0]){
                int ll = loc[lur][0];
                int tmp = g[ll - 1][k - 1];
                loc[lur][0] = ll - 1;
                loc[lur][1] = k - 1;
                loc[tmp][0] = ll;
                loc[tmp][1] = 0;
                swap(g[ll - 1][k - 1], g[ll][0]);
            }
        }
        else{
            int r1 = loc[lur][0];
            int r2 = loc[lur][1];
            int tmp = g[r1][r2 - 1];
            loc[lur][1] = r2 - 1;
            loc[tmp][1] = r2;
            swap(g[r1][r2 - 1], g[r1][r2]);
        }
    }
    printf("%I64u\n", ans);
    return 0;
}