hdu 5475(打破固定思维OR线段树)

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1327    Accepted Submission(s): 624

Problem Description

One day, a useless calculator was being built by Kuros. Let's assume that number X is showed on the screen of calculator. At first, X = 1. This calculator only supports two types of operation.
1. multiply X with a number.
2. divide X with a number which was multiplied before.
After each operation, please output the number X modulo M.

 

Input

The first line is an integer T(1≤T≤10), indicating the number of test cases.
For each test case, the first line are two integers Q and M. Q is the number of operations and M is described above. (1≤Q≤105,1≤M≤109)
The next Q lines, each line starts with an integer x indicating the type of operation.
if x is 1, an integer y is given, indicating the number to multiply. (0<y≤109)
if x is 2, an integer n is given. The calculator will divide the number which is multiplied in the nth operation. (the nth operation must be a type 1 operation.)

It's guaranteed that in type 2 operation, there won't be two same n.

 

Output

For each test case, the first line, please output "Case #x:" and x is the id of the test cases starting from 1.
Then Q lines follow, each line please output an answer showed by the calculator.

 

Sample Input

1 10 1000000000 1 2 2 1 1 2 1 10 2 3 2 4 1 6 1 7 1 12 2 7

 

Sample Output

Case #1: 2 1 2 20 10 1 6 42 504 84

 

Source

2015 ACM/ICPC Asia Regional Shanghai Online

 

题意:

初始化 s = 1

操作1 x,将 s = s*x%mod;

操作2 y,将第y步的x拿出来, s = s/x%mod;

问每一步操作得到的数字是多少。

开始看的时候完全不知道如何下手,有除法操作的取模运算,逆元？不行。然后找题解，然后发现自己的思维太局限了。除法不行的话那我们就将之前的的乘法操作取消就OK了啊！所以弄个标记数组，每一次出现除法的时候再对前前面的数扫一遍就行了，标记我们要除的那位，就等于之前没有乘过了。不过这样做很冒险的，O(n^2) 10^5的数据量。。然后就是强大的线段树的用场了。

4000ms+:

#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long LL;
int a[100005],vis[100005];
int main()
{
    int tcase;
    scanf("%d",&tcase);
    int t = 1;
    while(tcase--){
        printf("Case #%d:
",t++);
        int n;
        LL mod;
        LL s=1;
        scanf("%d%lld",&n,&mod);
        for(int i=1;i<=n;i++){
            vis[i] = false;
            int k,b;
            scanf("%d%d",&k,&b);
            if(k==1){
                vis[i] = true;
                a[i] = b;
                s=s*a[i]%mod;
            }else{
                s = 1;
                for(int j=1;j<i;j++){
                    if((LL)j==b){
                        vis[j] = false;
                    }else if(vis[j]){
                        s=s*a[j]%mod;
                    }
                }
            }
            printf("%lld
",s);
        }
    }
    return 0;
}

线段树版本:

1400ms+

#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long LL;
LL mod;
struct Tree{
    int l,r;
    LL v;
}tree[4*100005];
void pushup(int idx){
    tree[idx].v = (tree[idx<<1].v*tree[idx<<1|1].v)%mod;
}
void build(int l,int r,int idx){
    tree[idx].l = l;
    tree[idx].r = r;
    if(l==r){
        tree[idx].v = 1;
        return ;
    }
    int mid = (l+r)>>1;
    build(l,mid,idx<<1);
    build(mid+1,r,idx<<1|1);
    pushup(idx);
}
void update(int idx,int v,int id){
    if(tree[idx].l==tree[idx].r){
        tree[idx].v = v;
        return ;
    }
    int mid = (tree[idx].l+tree[idx].r)>>1;
    if(mid>=id) update(idx<<1,v,id);
    else update(idx<<1|1,v,id);
    pushup(idx);
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    int t = 1;
    while(tcase--){
        printf("Case #%d:
",t++);
        int n;
        LL s=1;
        scanf("%d%lld",&n,&mod);
        build(1,n,1);
        for(int i=1;i<=n;i++){
            int k,b;
            scanf("%d%d",&k,&b);
            if(k==1){
                update(1,b,i);
            }else{
                update(1,1,b);
            }
            printf("%lld
",tree[1].v);
        }
    }
    return 0;
}