CodeForces 438D The Child and Sequence
题目链接:http://codeforces.com/problemset/problem/438/D
The Child and Sequence
At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.
Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:
- Print operation l, r. Picks should write down the value of .
- Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r).
- Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x).
Can you help Picks to perform the whole sequence of operations?
Input
The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.
Each of the next m lines begins with a number type .
- If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1.
- If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2.
- If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3.
Output
For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.
Examples
5 5 1 2 3 4 5 2 3 5 4 3 3 5 1 2 5 2 1 3 3 1 1 3
8 5
10 10 6 9 6 7 6 1 10 10 9 5 1 3 9 2 7 10 9 2 5 10 8 1 4 7 3 3 7 2 7 9 9 1 2 4 1 6 6 1 5 9 3 1 10
49 15 23 1 9
Note
Consider the first testcase:
- At first, a = {1, 2, 3, 4, 5}.
- After operation 1, a = {1, 2, 3, 0, 1}.
- After operation 2, a = {1, 2, 5, 0, 1}.
- At operation 3, 2 + 5 + 0 + 1 = 8.
- After operation 4, a = {1, 2, 2, 0, 1}.
- At operation 5, 1 + 2 + 2 = 5.
思路:稍微分析一下题目,就能够看出这道题是线段树的单点更新,区间更新,区间求和问题。但是有一个问题存在,这里区间更新是取模,直接对和进行取模肯定是错误的,所以得换种想法做。考虑题目给的时间很充足,而且测试数据是单个样例测试,所以可以尝试区间更新时更新到叶子节点。但这种做法好像还是不太保险,仔细斟酌一下,其实对于取模操作,只有当该数大于模数,才需要更新,否则不需要,这样就类似于DFS的剪枝了,所以将每个区间的最大值求出来,更新时与模数比较,模数大就直接返回,不必向下更新了,以节约时间开销,这样就更有胜算了,事实证明是的,后来算了一下复杂度,发现这样做的话复杂度并不是很高。详见代码。
附上AC代码:
#include <bits/stdc++.h>
#define lrt rt<<1
#define rrt rt<<1|1
#define lson l, m, lrt
#define rson m+1, r, rrt
using namespace std;
typedef long long ll;
const int maxn = 100005;
ll sumv[maxn<<2], maxv[maxn<<2];
int n, q;void push_up(int rt){sumv[rt] = sumv[lrt]+sumv[rrt];maxv[rt] = max(maxv[lrt], maxv[rrt]);
}void build(int l, int r, int rt){if (l == r){int num;cin >> num;sumv[rt] = num;maxv[rt] = num;return ;}int m = (l+r)>>1;build(lson);build(rson);push_up(rt);
}void update1(int p, int val, int l, int r, int rt){if (l == r){sumv[rt] = val;maxv[rt] = val;return ;}int m = (l+r)>>1;if (p <= m)update1(p, val, lson);elseupdate1(p, val, rson);push_up(rt);
}void update2(int cl, int cr, int val, int l, int r, int rt){if (maxv[rt] < val)return ;if (l == r){sumv[rt] %= val;maxv[rt] %= val;return ;}int m = (l+r)>>1;if (cl <= m)update2(cl, cr, val, lson);if (cr > m)update2(cl, cr, val, rson);push_up(rt);
}ll query(int ql, int qr, int l, int r, int rt){if (ql<=l && r<=qr)return sumv[rt];int m = (l+r)>>1;ll sumr = 0;if (ql <= m)sumr += query(ql, qr, lson);if (qr > m)sumr += query(ql, qr, rson);return sumr;
}int main(){ios::sync_with_stdio(false);cin.tie(0);cin >> n >> q;build(1, n, 1);int type, l, r, x;while (q--){cin >> type >> l >> r;if (1 == type)cout << query(l, r, 1, n, 1) << endl;else if (2 == type){cin >> x;update2(l, r, x, 1, n, 1);}elseupdate1(l, r, 1, n, 1);}return 0;
}
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