CodeForces 438D The Child and Sequence（线段树）

清疚 2021-11-01 17:58 247阅读 0赞

题目链接：[http://codeforces.com/problemset/problem/438/D][http_codeforces.com_problemset_problem_438_D]  
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  
Input  
5 5  
1 2 3 4 5  
2 3 5 4  
3 3 5  
1 2 5  
2 1 3 3  
1 1 3  
Output  
8  
5  
Input  
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  
Output  
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.

分析：  
题目要求：单点更新，区间求和，区间更新。  
对于前两个来说，我们是很容易维护的。但是第三个是维护a\[i\]%x。这个值应该怎么维护呢？暴力维护？ 看到题目的时限还算长，又是单组测试。也许可以，不过不保险。  
对于这种情况，我们可以在结点中在维护一个结点最大值，这样如果x大于结点最大值，那么他继续搜下去，也就没有什么意义了。这个操作有点类似dfs的减枝把。

#include"stdio.h"
    #include"string.h"
    #include"algorithm"
    using namespace std;
    typedef long long ll;
    const int N = 100010;
    
    ll sum[N << 2];
    int max_val[N << 2],val[N << 2];
    int T,a[N],n,m;
    
    void Push_down(int id)
    {
        sum[id] = sum[id << 1] + sum[id << 1 | 1];
        max_val[id] = max(max_val[id << 1],max_val[id << 1 | 1]);
        return ;
    }
    void Build_Tree(int id,int l,int r)
    {
        sum[id] = 0; max_val[id] = -1;
        if(l == r)
        {
            val[id] = a[l];
            sum[id] = a[l]; max_val[id] = a[l];
            return ;
        }
        int mid = (l + r) >> 1;
        Build_Tree(id << 1,l,mid);
        Build_Tree(id << 1 | 1,mid + 1,r);
        Push_down(id);
    }
    
    ll Query_sum(int id,int L,int R,int l,int r)
    {
        if(l <= L && r >= R) return sum[id];
        int mid = (L + R) >> 1;
        ll ans = 0;
        if(l <= mid) ans += Query_sum(id << 1,L,mid,l,r);
        if(r > mid) ans += Query_sum(id << 1 | 1,mid + 1,R,l,r);
        return ans;
    }
    
    void Update(int id,int L,int R,int x,int v)
    {
        if(L == R)
        {
            max_val[id] = v; sum[id] = v; val[id] = v;
            return ;
        }
        int mid = (L + R) >> 1;
        if(x <= mid) Update(id << 1,L,mid,x,v);
        else Update(id << 1 | 1,mid + 1,R,x,v);
        Push_down(id);
    }
    
    void Update_query(int id,int L,int R,int l,int r,int v)
    {
        if(v > max_val[id]) return ;
        if(L == R)
        {
            val[id] = val[id] % v;
            max_val[id] = val[id];
            sum[id] = val[id]; return ;
        }
        int mid = (L + R) >> 1;
        if(l <= mid) Update_query(id << 1,L,mid,l,r,v);
        if(r > mid) Update_query(id << 1 | 1,mid + 1,R,l,r,v);
        Push_down(id);
    }
    
    int main()
    {
        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; i ++)
           scanf("%d",&a[i]);
        Build_Tree(1,1,n);
        for(int i = 1; i <= m;i ++)
        {
            int op,x,y;
            scanf("%d%d%d",&op,&x,&y);
            if(op == 1)
            {
                printf("%lld\n",Query_sum(1,1,n,x,y));
            }
            if(op == 2)
            {
                int t; scanf("%d",&t);
                Update_query(1,1,n,x,y,t);
            }
            if(op == 3)
            {
                Update(1,1,n,x,y);
            }
        }
    }

[http_codeforces.com_problemset_problem_438_D]: http://codeforces.com/problemset/problem/438/D