hdu 3074(线段树)

最新推荐文章于 2025-12-17 19:02:04 发布

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原文链接：http://www.cnblogs.com/liyinggang/p/5531237.html

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#java

本文介绍了一款名为“乘法游戏”的编程挑战，玩家需要在数字序列中进行数值替换并计算子序列的乘积，采用线段树数据结构优化算法效率。

Multiply game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2211 Accepted Submission(s): 794

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

线段树一直交给队友弄，，好久没练过，自己手打错了好多次，，看来还是要多练练。。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 50005;
const LL mod =1000000007;
LL a[N];
struct Tree{
    int l,r,mid;
    LL v;
}tree[N<<2];
void build(int l,int r,int idx){
    tree[idx].l = l;
    tree[idx].r = r;
    tree[idx].mid = (l+r)>>1;
    if(l==r){
        tree[idx].v=a[l];
        return;
    }
    build(l,tree[idx].mid,idx<<1);
    build(tree[idx].mid+1,r,idx<<1|1);
    tree[idx].v =(tree[idx<<1].v%mod)*(tree[idx<<1|1].v%mod)%mod;
}
void update(int pos,LL v,int idx){
    if(tree[idx].l==tree[idx].r){
        tree[idx].v = v;
        return;
    }
    if(pos<=tree[idx].mid) update(pos,v,idx<<1);
    else update(pos,v,idx<<1|1);
    tree[idx].v =(tree[idx<<1].v%mod)*(tree[idx<<1|1].v%mod)%mod;
}
LL query(int l,int r,int idx){
    if(tree[idx].l==l&&tree[idx].r==r){
        return tree[idx].v%mod;
    }
    if(r<=tree[idx].mid) return query(l,r,idx<<1)%mod;
    else if(l>tree[idx].mid) return query(l,r,idx<<1|1)%mod;
    else return (query(l,tree[idx].mid,idx<<1)%mod)*(query(tree[idx].mid+1,r,idx<<1|1)%mod)%mod;
}
int main()
{
    int n;
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){scanf("%lld",&a[i]);}
        build(1,n,1);
        int Q;
        scanf("%d",&Q);
        while(Q--){
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            if(a==0){
                query(b,c,1);
                printf("%lld\n",query(b,c,1));
            }else update(b,(LL)c,1);
        }
    }
    return 0;
}
 