CF438D The Child and Sequence

题意大意

给定数列，区间查询和，区间取模，单点修改。

(n),(m)小于(10^5)

题目描述

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 mm 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?

输入输出格式

输入格式：

The first line of input contains two integer: (n,m) ((1<=n,m<=10^{5})) . The second line contains nnintegers, separated by space: (a[1],a[2],...,a[n] (1<=a[i]<=10^{9})) — initial value of array elements.

Each of the next mm lines begins with a number typetype .

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<=10^{9})) , which correspond the operation (2).
If (type=3), there will be two integers more in the line: (k,x (1<=k<=n; 1<=x<=10^{9})) , which correspond the operation (3).

输出格式：

For each operation (1), please print a line containing the answer. Notice that the answer may exceed the (32)-bit integer.

输入输出样例

输入样例#1：

输出样例#1：

8
5

输入样例#2：

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

输出样例#2：

说明

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：

题目大意：给定数列，区间查询和，区间取模，单点修改。

(n,m)小于(10^5)

查询区间和和单点修改就不用说了吧，线段树基本操作，那？？对于这道题目的区间修改该怎么处理呢？？不难发现，一个数模两次同样的数时没有任何意义的，而且一个数模一个比它大的数也是没有意义的？！！！那么，我们便可以采用一种暴力的思想，去修改每一个位置，每次判断一下区间最大值是不是比模数大即可。因为是暴力修改，所以也无需(pushdown)，那么代码应该就很好写了吧，也不算太长。

自己整理的题解

下面是我的代码，仅供参考，~~毕竟太丑了~~：

#include<cstdio>
#include<algorithm>
#include<cctype>
#define maxn 100007
#define ls rt<<1
#define rs rt<<1|1
#define ll long long
using namespace std;
int n,m,maxx[maxn<<2];
ll sum[maxn<<2];
inline int qread() {
  char c=getchar();int num=0,f=1;
  for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
  for(;isdigit(c);c=getchar()) num=num*10+c-'0';
  return num*f;
}
inline void pushup(int rt) {
  sum[rt]=sum[ls]+sum[rs];
  maxx[rt]=max(maxx[ls],maxx[rs]);
}
void build(int rt, int l, int r) {
  if(l==r) {
    maxx[rt]=sum[rt]=qread();
    return;
  }
  int mid=(l+r)>>1;
  build(ls,l,mid);
  build(rs,mid+1,r);
  pushup(rt);
}
void add(int rt, int l, int r, int L, int val) {
  if(l==r) {
    maxx[rt]=sum[rt]=val;
    return;
  }
  int mid=(l+r)>>1;
  if(L<=mid) add(ls,l,mid,L,val);
  else add(rs,mid+1,r,L,val);
  pushup(rt);
}
ll csum(int rt, int l, int r, int L, int R) {
  if(L<=l&&r<=R) return sum[rt];
  int mid=(l+r)>>1;
  ll ans=0;
  if(L<=mid) ans+=csum(ls,l,mid,L,R);
  if(R>mid) ans+=csum(rs,mid+1,r,L,R);
  return ans;
}
void modify(int rt, int l, int r, int L, int R, int p) {
  if(maxx[rt]<p) return;
  if(l==r) {
    sum[rt]%=p;maxx[rt]%=p;
    return;
  }
  int mid=(l+r)>>1;
  if(L<=mid) modify(ls,l,mid,L,R,p);
  if(R>mid) modify(rs,mid+1,r,L,R,p);
  pushup(rt);
}
int main() {
  n=qread(),m=qread();
  build(1,1,n);
  for(int i=1,k,x,y,z;i<=m;++i) {
    k=qread(),x=qread(),y=qread();
    if(k==1) printf("%lld
",csum(1,1,n,x,y));
    if(k==2) {
      z=qread();
      modify(1,1,n,x,y,z);
    }
    if(k==3) add(1,1,n,x,y);
  }
  return 0;
}