HDU 3074.Multiply game-区间乘法-线段树(单点更新、区间查询)，上推标记取模

佚名 6年前 (2019-04-09) 随笔 786人围观抢沙发百度已收录

Multiply game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3224 Accepted Submission(s): 1173

Problem Description Tired of playing computer games, alpc23 is planning to play a game on numbers. Because plus and subtraction is too easy for this gay, he wants to do some multiplication in a number sequence. After playing it a few times, he has found it is also too boring. So he plan to do a more challenge job: he wants to change several numbers in this sequence and also work out the multiplication of all the number in a subsequence of the whole sequence.
To be a friend of this gay, you have been invented by him to play this interesting game with him. Of course, you need to work out the answers faster than him to get a free lunch, He he…

SRE实战互联网时代守护先锋，助力企业售后服务体系运筹帷幄！一键直达领取阿里云限量特价优惠。 Input The first line is the number of case T (T<=10).
For each test case, the first line is the length of sequence n (n<=50000), the second line has n numbers, they are the initial n numbers of the sequence a1,a2, …,an,
Then the third line is the number of operation q (q<=50000), from the fourth line to the q+3 line are the description of the q operations. They are the one of the two forms:
0 k1 k2; you need to work out the multiplication of the subsequence from k1 to k2, inclusive. (1<=k1<=k2<=n)
1 k p; the kth number of the sequence has been change to p. (1<=k<=n)
You can assume that all the numbers before and after the replacement are no larger than 1 million.

Output For each of the first operation, you need to output the answer of multiplication in each line, because the answer can be very large, so can only output the answer after mod 1000000007.

Sample Input 1 6 1 2 4 5 6 3 3 0 2 5 1 3 7 0 2 5

Sample Output 240 420

Source 2009 Multi-University Training Contest 17 - Host by NUDT

没什么好说的，水题。

代码:

 1 //HDU 3074.Multiply game-区间乘法-线段树(单点更新+区间查询)
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 typedef long long ll;
 5 const int maxn=5e5+10;
 6 const ll mod=1000000007;
 7 #define lson l,m,rt<<1
 8 #define rson m+1,r,rt<<1|1
 9 
10 ll tree[maxn<<2];
11 
12 ll pushup(int rt)
13 {
14     tree[rt]=(tree[rt<<1]*tree[rt<<1|1])%mod;
15 }
16 
17 void build(int l,int r,int rt)
18 {
19     if(l==r){
20         scanf("%lld",&tree[rt]);
21         return ;
22     }
23 
24     int m=(l+r)>>1;
25     build(lson);
26     build(rson);
27     pushup(rt);
28 }
29 
30 void update(int pos,ll c,int l,int r,int rt)
31 {
32     if(l==r){
33         tree[rt]=c;
34         return ;
35     }
36 
37     int m=(l+r)>>1;
38     if(pos<=m) update(pos,c,lson);
39     if(pos> m) update(pos,c,rson);
40     pushup(rt);
41 }
42 
43 ll query(int L,int R,int l,int r,int rt)
44 {
45     if(L>r||l>R) return 0;
46     if(L<=l&&r<=R){
47         return tree[rt];
48     }
49 
50     int m=(l+r)>>1;
51     ll ret=1;
52     if(L<=m) ret=(ret*query(L,R,lson))%mod;
53     if(R> m) ret=(ret*query(L,R,rson))%mod;
54     return ret;
55 }
56 
57 int main()
58 {
59     int t;
60     scanf("%d",&t);
61     while(t--){
62         int n;
63         scanf("%d",&n);
64         build(1,n,1);
65         int m;
66         scanf("%d",&m);
67         for(int i=1;i<=m;i++){
68             int op;
69             scanf("%d",&op);
70             if(op==0){
71                 int l,r;
72                 scanf("%d%d",&l,&r);
73                 printf("%lld\n",query(l,r,1,n,1));
74             }
75             else{
76                 int pos;ll val;
77                 scanf("%d%lld",&pos,&val);
78                 val=val%mod;
79                 update(pos,val,1,n,1);
80             }
81         }
82     }
83 }