461C. Appleman and a Sheet of Paper(树状数组）

C. Appleman and a Sheet of Paper

time limit per test

2 seconds

memory limit per test

256 megabytes

Appleman has a very big sheet of paper. This sheet has a form of rectangle with dimensions 1 × n. Your task is help Appleman with folding of such a sheet. Actually, you need to perform q queries. Each query will have one of the following types:

1. Fold the sheet of paper at position p_i. After this query the leftmost part of the paper with dimensions 1 × p_i must be above the rightmost part of the paper with dimensions 1 × ([current width of sheet] - p_i).
2. Count what is the total width of the paper pieces, if we will make two described later cuts and consider only the pieces between the cuts. We will make one cut at distance l_i from the left border of the current sheet of paper and the other at distance r_i from the left border of the current sheet of paper.

Please look at the explanation of the first test example for better understanding of the problem.

Input

The first line contains two integers: n and q (1  ≤ n ≤ 10⁵; 1 ≤ q ≤ 10⁵) — the width of the paper and the number of queries.

Each of the following q lines contains one of the described queries in the following format:

• "1 p_i" (1 ≤ p_i < [current width of sheet]) — the first type query.
• "2 l_i r_i" (0 ≤ l_i < r_i ≤ [current width of sheet]) — the second type query.

Output

For each query of the second type, output the answer.

Sample test(s)

input

7 4
1 3
1 2
2 0 1
2 1 2

output

4
3

input

10 9
2 2 9
1 1
2 0 1
1 8
2 0 8
1 2
2 1 3
1 4
2 2 4

output

7
2
10
4
5

 

题意：给你一张1*n的纸片，有2种操作：（1）1 x，在x的地方将左半部分向左边折叠。（2）2 x y 求[x,y]区间的纸片厚度总和

思路：其实是一道模拟题，因为N的范围，和多次求区间和，明显应该用树状数组。。在折叠的过程中折叠后的纸片会变小。在1—N的线段上覆盖的长度也在减小，每次折叠完会多出一些没用的区间（想象一下不用将纸片前移到0，只是不断的折叠），所以当折叠过去后超过当前的右边界。。可以等效为将右边的向左折，相当于翻转了一次。。(不能向左折因为超出的部分会加上没用的区间的厚度）下次再进行折叠操作的时候就看成起始端在右边。。你可以设置标记。。2次翻转就恢复正常样子

代码：（参考别人的）：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
const int maxn = 100005;
int bit[maxn];
int n, q;
int lowbit(int x)
{
    return x & -x;
}
void add(int x, int y)
{
    while(x <= n)
    {
        bit[x] += y;
        x += lowbit(x);
    }
}
int getsum(int x)
{
    int ans = 0;
    while(x > 0)
    {
        ans += bit[x];
        x -= lowbit(x);
    }
    return ans;
}
int main()
{
    int t, p, x, y;
    scanf("%d%d", &n, &q);
    memset(bit, 0, sizeof(bit));
    for(int i = 1; i <= n; i++) add(i, 1);
    bool flag = 0;
    int l = 1, r = n;
    for(int i = 0; i < q; i++)
    {
        scanf("%d", &t);
        if(t == 1)
        {
            scanf("%d", &p);
            if(!flag) p = l + p;//根据纸片是否与正常翻转过来重新计算折纸点
            else p = r - p + 1;
            if(2 * p > r + l + 1)//右半部分小，向左边叠
            {
                for(int j = p; j <= r; j++)
                {
                    int u = getsum(j) - getsum(j - 1);//下标从0开始对树状数组处理时都要减1.下同
                    add(2 * p - j- 1, u);
                }
                r = p - 1;
                if(flag == 0) flag ^= 1;
            }
            else
            {
                for(int j = l; j < p; j++)
                {
                    int u = getsum(j) - getsum(j - 1);
                    add(2 * p - 1 - j, u);
                }
                l = p;
                if(flag == 1) flag ^= 1;
            }
        }
        else
        {
            scanf("%d%d", &x, &y);
            if(!flag)
            {
                x = l + x - 1;
                y = l + y - 1;
            }
            else
            {
                int t = x;
                x = r - y;
                y = r - t;
            }
            printf("%d\n", getsum(y) - getsum(x));
        }
    }
    return 0;
}

 

下面一种类似，还省去了标记：只用看L与R 的关系判断上次有没有翻转过

#include <cstdio>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define REP(i,n) for((i)=0;(i)<(int)(n);(i)++)
int tree[1<<20];
void add(int pos, int val)
{
    for (int i = pos; i<(1<<20); i = ((i) | (i + 1))) tree[i] += val;
}
int sum(int pos)
{
    int ans = 0;
    for (int i = pos; i>0; i = ((i)&(i - 1))) ans += tree[i - 1];
    return ans;
}
int get(int x)
{
    return sum(x + 1) - sum(x);
}
void update(int x, int val)
{
    add(x, val);
}
int sum(int L, int R)
{
    return sum(R) - sum(L);
}
int main()
{
    int N, Q, i, j;
    scanf("%d%d", &N, &Q);
    int L = 0, R = N;
    REP(i, N) update(i, 1);
    REP(j, Q)
    {
        int type;
        scanf("%d", &type);
        if (type == 1)
        {
            int p;
            scanf("%d", &p);
            int M;
            if (L < R) M = L + p;
            else M = L - p;
            if (L > R) swap(L, R);
            int w = min(M - L, R - M);
            int L2, R2;
            if (M - L <= R - M)
            {
                REP(i, w)
                {
                    int tmp = get(M - 1 - i);
                    update(M + i, tmp);
                }
                L2 = M;
                R2 = R;
            }
            else
            {
                REP(i, w)
                {
                    int tmp = get(M + i);
                    update(M - 1 - i, tmp);
                }
                L2 = M;
                R2 = L;
            }
            L = L2;
            R = R2;
        }
        else
        {
            int l, r;
            scanf("%d%d", &l, &r);
            int ans = 0;
            if (L < R) ans = sum(L + l, L + r);
            else ans = sum(L - r, L - l);
            printf("%d\n", ans);
        }
    }
}