hdu_3074Multiply game线段树+点修改

Multiply game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1501    Accepted Submission(s): 521

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…

 

 

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

 

注意此题数值应设为 long long

#include<stdio.h>
#include<cstdio>
#include <cstring>
#include<iostream>
#include<algorithm>
#define inf -0x3f3f3f3f
#define INF 0x3f3f3f3f
#define mem0(a) memset(a,0,sizeof(a))
#define mod 1000000007
using namespace std;
const int  num = 50000+10;
long long  a[num<<2];
void GetMul(int cur){
    a[cur] = (a[cur<<1]*a[cur<<1|1])%mod;
}
void update(int p ,int v,int l,int r,int cur){
    if(l == r){
        a[cur] = v;//叶结点，直接更新a
    }
    else {
    int mid = (r + l )>>1 ;
        if(p <= mid ){//递归更新左子树和右子树
            update(p,v,l,mid,cur<<1);
        }
        else {
            update(p,v,mid+1,r,cur<<1|1);
        }
        GetMul(cur);//计算本结点的a
    }
}
void build(int l,int r,int cur){
    if( l == r ){
        scanf("%d",&a[cur]);
        return ;
    }
    int  mid= (l + r )>>1;
    build(l,mid,cur<<1);
    build(mid+1,r,cur<<1|1);
    GetMul(cur);
}

long long   query(int ql,int qr,int l,int r,int cur){
    int mid = ( l + r )>>1 ;
  long long  ans = 1;
    if(ql <= l && qr >= r)  return a[cur];
    if(ql <= mid)
        ans =(ans * query(ql,qr,l,mid,cur<<1))%mod;
    if(qr > mid )
        ans =(ans * query(ql,qr,mid+1,r,cur<<1|1))%mod;
    return ans  ;
}
int main()
{
    int N,M,T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        build(1,N,1);
        scanf("%d",&M);
        while(M--){
            int a,b,c;
            scanf("%d%d%d",&c,&a,&b);
            if( c == 0 )
                printf("%lld\n",query(a,b,1,N,1));
            else if( c == 1)
                update(a,b,1,N,1);
        }
    }
    return 0;
}