hdu 5475 An easy problem(线段树)

最新推荐文章于 2019-08-12 15:28:34 发布

原创最新推荐文章于 2019-08-12 15:28:34 发布 · 170 阅读

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本文介绍了一个简单的计算器模拟问题，该计算器支持两种操作：乘法和特定的除法，并需要输出每次操作后屏幕上的数字模M的结果。通过构建线段树进行区间更新和查询，实现了高效地解决此问题。

An easy problem

Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2826 Accepted Submission(s): 1056

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

Sample Output

思路：
看成是求n个数的乘积，op==1,把对应点的数字改为x
op==2， 把x位置的数改为1。每次输出1~n的乘积
代码：
#include<bits/stdc++.h>
using namespace std;
#define ll long long 
const int maxn=100005;
ll mod,tree[maxn<<2],X[maxn],K[maxn];
void build(int l,int r,int rt)
{
    if(l==r)
    {
        tree[rt]=1%mod;
        return;
    }
    int mid=(l+r)/2;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    tree[rt]=tree[rt<<1]*tree[rt<<1|1]%mod;
}
void update(int L,ll C,int l,int r,int rt)
{
    if(l==r)
    {
        tree[rt]=C%mod;
        return;
    }
    int mid=(l+r)/2;
    if(L<=mid)update(L,C,l,mid,rt<<1);
    else update(L,C,mid+1,r,rt<<1|1);
    tree[rt]=tree[rt<<1]*tree[rt<<1|1]%mod;
}
int main()
{
    int T;scanf("%d",&T);
    for(int i=1;i<=T;i++)
    {
        printf("Case #%d:\n",i);
        int Q;scanf("%d%lld",&Q,&mod);
        build(1,100000,1);
        for(int i=1;i<=Q;i++)
        {
            int op;ll x;scanf("%d%lld",&op,&x);
            if(op==1) 
            {
                update(i,x,1,100000,1);
                printf("%lld\n",tree[1]);
            }
            else
            {
                update(x,1,1,100000,1);
                printf("%lld\n",tree[1]);
            }
        }
    }
    return 0;
}