首页 > ACM题库 > HDU-杭电 > HDU 3074-Multiply game-线段树-[解题报告]HOJ
2014
03-01

HDU 3074-Multiply game-线段树-[解题报告]HOJ

Multiply game

问题描述 :

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…

输入:

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.

输出:

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.

样例输入:

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

样例输出:

240
420

#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
#include <cstdio>
#define L(rt) (rt<<1)
#define R(rt)  (rt<<1|1)
using namespace std;

const int mod=1000000007;
const int N=50005;
int  num[N];

struct node
{
    int l,r;
    long long ans;
}tree[N*4];

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("1.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("%lld\n",ans%mod);
            }
            else
            update(y,x,1);
        }          
    }
    return 0;
}

参考:http://blog.csdn.net/weyuli/article/details/9383195


  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  2. 很高兴你会喜欢这个网站。目前还没有一个开发团队,网站是我一个人在维护,都是用的开源系统,也没有太多需要开发的部分,主要是内容整理。非常感谢你的关注。

  3. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮