Multiply game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1190 Accepted Submission(s): 394
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…
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.
#include <iostream>
using namespace std;
#define lson l,m,rt*2
#define rson m+1,r,rt*2+1
#define maxn 50005
__int64 sum[maxn<<2];
void build(int l,int r,int rt)
{
if(l==r)
{
scanf("%I64d",&sum[rt]);
return;
}
int m=(r+l)/2;
build(lson);
build(rson);
sum[rt]=(sum[rt*2]*sum[rt*2+1])%1000000007;
}
void update(int p,int x,int l,int r,int rt)
{
if (l==r)
{
sum[rt]=x;
return;
}
int m=(r+l)/2;
if(p<=m)
update(p,x,lson);
else
update(p,x,rson);
sum[rt]=(sum[rt*2]*sum[rt*2+1])%1000000007;
}
__int64 query(int L,int R,int l,int r,int rt)
{
if (l>=L&&r<=R)
return sum[rt]%1000000007;
int m=(r+l)/2;
__int64 ret=1;
if (L<=m)
ret*=query(L,R,lson)%1000000007;
if(R>m)
ret*=query(L,R,rson)%1000000007;
return ret%1000000007;
}
int main()
{
int t,n,q,a,b,c;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
build(1,n,1);
scanf("%d",&q);
while(q--)
{
scanf("%d%d%d",&a,&b,&c);
if(a==0)
printf("%I64d\n",query(b,c,1,n,1));
else
if (a==1)
update(b,c,1,n,1);
}
}
return 0;
}
切記用64位,否則WA!