CF 292E, 线段树

题目大意：给你a，b两个数组，两个操作，一个是把a里连续的一段复制到b中，另一个是单点查询b。

解：可以看到其实a是不修改的，本质我们要维护b到a的一个映射，而赋值又是连续的，所以本质又是维护公差1的等差序列，那么这个等差序列只要一个首项就能表达他自己的性质。所以我们每个线段存一个首项，lazytag维护即可。

#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <complex>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <deque>
#include <ctime>

using namespace std;

const double EPS = 1e-8;

#define ABS(x) ((x)<0?(-(x)):(x))
#define SQR(x) ((x)*(x))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define MAX(a,b) ((a)>(b)?(a):(b))

#define LSON(x) ((x)<<1)
#define RSON(x) (((x)<<1)+1)
#define LOWBIT(x) ((x)&(-(x)))
#define MAXS 1111
#define MAXN 222222
#define VOIDPOINT 0
#define LL long long
#define OO 214748364

int a[MAXN], b[MAXN];

struct segTree{
    int tag[MAXN*4], num[MAXN*4], l[MAXN*4], r[MAXN*4], mid[MAXN*4];
    int gx, gy, gz;
    inline void updata(const int &kok) {
    }
    inline void pushdown(int kok) {
        if (l[kok] == r[kok]) num[kok] = a[tag[kok]];
        else {
            tag[LSON(kok)] = tag[kok];
            tag[RSON(kok)] = tag[kok] + (r[LSON(kok)] - l[LSON(kok)] + 1);
        }
        tag[kok] = 0;
    }
    void build(int kok, int ll, int rr, int *a) {
        l[kok] = ll; r[kok] = rr; tag[kok] = 0;
        if (ll == rr) {
            num[kok] = a[ll]; 
            return ;
        }
        int m = mid[kok] = (ll + rr) >> 1;
        build(LSON(kok), ll, m, a); build(RSON(kok), m+1, rr, a);
    }
    void insert(int kok) {
//        cout << "insert ::  " << l[kok] << ' '<< r[kok] << endl;

        if (tag[kok]) pushdown(kok);

        if (gx <= l[kok] && r[kok] <= gy) {
            tag[kok] = gz + l[kok] - gx; return ;    
        }
        if (gx <= mid[kok]) insert(LSON(kok));
        if (gy > mid[kok]) insert(RSON(kok));
    }
    int query(int kok) {
        if (tag[kok]) pushdown(kok);

        if (l[kok] == r[kok]) {
            return num[kok];
        }
        if (gx <= mid[kok]) 
            return query(LSON(kok));
        else if (gx > mid[kok]) 
            return query(RSON(kok));
    }
    void set(int a = 0, int b = 0, int c = 0) {
        gx = a; gy = b; gz = c;
    }
} Tree;

int n, m;

int main() {
//    freopen("test.txt", "r", stdin);    
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= n; ++i) scanf("%d", &b[i]);
    int cas, x, y, k;
    Tree.build(1, 1, n, b);
    while (m--) {
        scanf("%d", &cas);
        if (cas == 1) {
            scanf("%d%d%d", &x, &y, &k);
            Tree.set(y, y+k-1, x);
            Tree.insert(1);
        } else {
            scanf("%d", &x);
            Tree.gx = x;
            printf("%d
", Tree.query(1));

        }
    }

    return 0;
}