Multiply game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1055 Accepted Submission(s): 343
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…
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.
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
Recommend
lcy
题意：两种操作
0 k1 k2; you need to work out the multiplication of the subsequence from k1 to k2, inclusive.
1 k p; the kth number of the sequence has been change to p. (1<=k<=n)
思路：线段树 跟前面做的几道对比 有点简单了 但要注意要用__int64 而且每乘完后都要取模
#include<iostream> #include<cstdio> #include<cstring> using namespace std; #define L(rt) (rt<<1) #define R(rt) (rt<<1|1) const int mod=1000000007; const int N=50010; int num[N]; struct node{ int l,r; long long ans; }tree[N*3]; void build(int l,int r,int rt){ tree[rt].l=l; tree[rt].r=r; if(l==r){ tree[rt].ans=num[l]; return ; } int mid=(l+r)>>1; build(l,mid,L(rt)); build(mid+1,r,R(rt)); tree[rt].ans=(tree[L(rt)].ans*tree[R(rt)].ans)%mod; } long long query(int l,int r,int rt){ if(tree[rt].l==l && tree[rt].r==r) return tree[rt].ans; int mid=(tree[rt].l+tree[rt].r)>>1; if(r<=mid) return query(l,r,L(rt)); else if(l>=mid+1) return query(l,r,R(rt)); else{ long long a=query(l,mid,L(rt)); long long b=query(mid+1,r,R(rt)); return (a*b)%mod; } } void update(int val,int loc,int rt){ if(tree[rt].l==loc && tree[rt].r==loc){ tree[rt].ans=val; return ; } if(loc<=tree[L(rt)].r) update(val,loc,L(rt)); if(loc>=tree[R(rt)].l) update(val,loc,R(rt)); tree[rt].ans=(tree[L(rt)].ans*tree[R(rt)].ans)%mod; } int main(){ //freopen("input.txt","r",stdin); int t,n,m; scanf("%d",&t); while(t--){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&num[i]); build(1,n,1); scanf("%d",&m); int op,x,y; while(m--){ scanf("%d%d%d",&op,&x,&y); if(op==0){ long long ans=query(x,y,1); printf("%I64d\n",ans%mod); }else update(y,x,1); } } return 0; }