CodeForces 438D The Child and Sequence（线段树）
题目链接：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);}}}
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左手的ㄟ右手
我就是我
爱被打了一巴掌
╰+哭是因爲堅強的太久メ
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