CodeForces 438D 线段树剪枝

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 $\text{[math]}$ .
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 $\text{[math]}$ .

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.

Sample test(s)

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.

注意：剪枝：若x<mod,x=x;

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 
 10 #define N 100010
 11 #define M 15
 12 //#define mod 1000000007
 13 //#define mod2 100000000
 14 #define ll long long
 15 #define maxi(a,b) (a)>(b)? (a) : (b)
 16 #define mini(a,b) (a)<(b)? (a) : (b)
 17 
 18 using namespace std;
 19 
 20 int n,q;
 21 int x[N];
 22 int type;
 23 int a,b,c;
 24 
 25 struct node
 26 {
 27     ll sum;
 28     int ma;
 29 }tree[4*N];
 30 /*
 31 void update(int l,int r,int i)
 32 {
 33     if(!tree[i].mark) return;
 34     int mid=(l+r)/2;
 35     tree[2*i].sum+=tree[i].mark*(ll)(mid-l+1);
 36     tree[2*i+1].sum+=tree[i].mark*(ll)(r-mid);
 37     tree[2*i].mark+=tree[i].mark;
 38     tree[2*i+1].mark+=tree[i].mark;
 39     tree[i].mark=0;
 40 
 41 }*/
 42 
 43 ll query(int tl,int tr,int l,int r,int i)
 44 {
 45     if(tl>r || tr<l)
 46         return 0;
 47     if(tl<=l && r<=tr)
 48         return tree[i].sum;
 49    // update(l,r,i);
 50     int mid=(l+r)/2;
 51 
 52     return query(tl,tr,l,mid,2*i)+query(tl,tr,mid+1,r,2*i+1);
 53 }
 54 /*
 55 void addd(int tl,int tr,int l,int r,int i,int val)
 56 {
 57     if(tl>r || tr<l)
 58         return;
 59     if(tl<=l && tr>=r){
 60         tree[i].sum+=val*(ll)(r-l+1);
 61         tree[i].mark+=val;
 62         return;
 63     }
 64     update(l,r,i);
 65     int mid=(l+r)/2;
 66     addd(tl,tr,l,mid,2*i,val);
 67     addd(tl,tr,mid+1,r,2*i+1,val);
 68     tree[i].sum=tree[2*i].sum+tree[2*i+1].sum;
 69 }*/
 70 
 71 void modefiyMod(int tl,int tr, int l, int r,int i,int mod)
 72 {
 73     if(tree[i].ma<mod) return;
 74     if(l==r){
 75         tree[i].sum=tree[i].ma=tree[i].sum%mod;
 76         return;
 77     }
 78     else{
 79         int mid=(l+r)/2;
 80         if(tr<=mid){
 81             modefiyMod(tl,tr,l,mid,i*2,mod);
 82         }
 83         else if(tl>mid){
 84             modefiyMod(tl,tr,mid+1,r,i*2+1,mod);
 85         }
 86         else{
 87             modefiyMod(tl,tr,l,mid,i*2,mod);
 88             modefiyMod(tl,tr,mid+1,r,i*2+1,mod);
 89         }
 90         tree[i].sum=tree[2*i].sum+tree[2*i+1].sum;
 91         tree[i].ma=max(tree[2*i].ma,tree[2*i+1].ma);
 92     }
 93 }
 94 
 95 void setVal(int i, int l, int r,int x,int v)
 96 {
 97     if(l==r){
 98         tree[i].sum=tree[i].ma=v;
 99         return;
100     }
101     else{
102         int mid=(l+r)/2;
103         if(x<=mid){
104             setVal(i*2,l,mid,x,v);
105         }
106         else{
107             setVal(i*2+1,mid+1,r,x,v);
108         }
109         tree[i].sum=tree[2*i].sum+tree[2*i+1].sum;
110         tree[i].ma=max(tree[2*i].ma,tree[2*i+1].ma);
111     }
112 }
113 
114 void build_tree(int l,int r,int i)
115 {
116     if(l==r){
117         tree[i].sum=x[l];
118         tree[i].ma=x[l];
119         return;
120     }
121     int mid=(l+r)/2;
122     build_tree(l,mid,2*i);
123     build_tree(mid+1,r,2*i+1);
124     tree[i].sum=tree[2*i].sum+tree[2*i+1].sum;
125     tree[i].ma=max(tree[2*i].ma,tree[2*i+1].ma);
126 }
127 
128 int main()
129 {
130     int i;
131     //freopen("data.in","r",stdin);
132     //scanf("%d",&T);
133     //for(int cnt=1;cnt<=T;cnt++)
134     //while(T--)
135     while(scanf("%d%d",&n,&q)!=EOF)
136     {
137        for(i=1;i<=n;i++) scanf("%d",&x[i]);
138        memset(tree,0,sizeof(tree));
139        build_tree(1,n,1);
140        while(q--){
141             scanf("%d",&type);
142             if(type==1){
143                 scanf("%d%d",&a,&b);
144                 ll ans=query(a,b,1,n,1);
145                 printf("%I64d\n",ans);
146 
147             }
148             else if(type==2){
149                 scanf("%d%d%d",&a,&b,&c);
150                 modefiyMod(a, b, 1, n, 1, c);
151             }
152             else{
153                 scanf("%d%d",&a,&b);
154                 setVal(1, 1, n, a, b);
155             }
156        }
157     }
158 
159     return 0;
160 }

posted on 2014-08-18 18:33 njczy2010 阅读(249) 评论(0) 编辑收藏举报

CodeForces 438D 线段树 剪枝

CodeForces 438D 线段树剪枝