标签：can multi its hang name use play efi pos
InputThe 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.
OutputFor 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
题意：输入n个数字，m次查询，每次查询三个数 a b c ，a=0时求b~c的乘积 a=1时将下标b的值修改为c
#include<bits/stdc++.h> using namespace std; typedef long long ll; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 const int maxn = 50000+10; const int INF = 0x3f3f3f3f; const int Mod = 1000000007; struct node { int l,r; ll sum; } tree[maxn<<2]; void pushup(int rt) { tree[rt].sum=(tree[rt<<1].sum*tree[rt<<1|1].sum)%Mod; } void build(int l,int r,int rt) { tree[rt].l=l; tree[rt].r=r; if(l==r) { scanf("%lld",&tree[rt].sum); return; } int m=(l+r)>>1; build(lson); build(rson); pushup(rt); } void updata(int rt,int pos, ll c) { if(tree[rt].l==tree[rt].r) { tree[rt].sum=c; return ; } int m=(tree[rt].l+tree[rt].r)>>1; if(pos<=m) updata(rt<<1,pos,c); else updata(rt<<1|1,pos,c); pushup(rt); } ll query(int l,int r,int rt) { if(tree[rt].l>=l&&tree[rt].r<=r) { return tree[rt].sum; } ll ans=1; int m=(tree[rt].l+tree[rt].r)>>1; if(l<=m) ans=ans*query(l,r,rt<<1)%Mod; if(r>m) ans=ans*query(l,r,rt<<1|1)%Mod; return ans; } int main() { int t,n,m; ll a,b,c; scanf("%d",&t); while(t--) { scanf("%d",&n); build(1,n,1); scanf("%d",&m); while(m--) { scanf("%lld %lld %lld",&a,&b,&c); if(a) { updata(1,b,c); } else printf("%I64d\n",query(b,c,1)); } } }
标签：can multi its hang name use play efi pos
原文地址：https://www.cnblogs.com/MengX/p/9265512.html
