首页 > 代码库 > hdu 5475 线段树

hdu 5475 线段树

2024-08-11 19:50:48   221人阅读

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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
110 10000000001 22 11 21 102 32 41 61 71 122 7
 

 

Sample Output
Case #1:2122010164250484
 

 

Source
2015 ACM/ICPC Asia Regional Shanghai Online
思路：线段树单点更新；
#include<bits/stdc++.h>using namespace std;#define ll long long#define pi (4*atan(1.0))const int N=1e5+10,M=4e6+10,inf=1e9+10;ll sum[N<<2],mod;void pushup(int pos){    sum[pos]=(sum[pos<<1|1]*sum[pos<<1])%mod;}void buildtree(int l,int r,int pos){    if(l==r)    {        sum[pos]=1;        return;    }    int mid=(l+r)>>1;    buildtree(l,mid,pos<<1);    buildtree(mid+1,r,pos<<1|1);    pushup(pos);}void update(int point,ll change,int l,int r,int pos){    if(l==r&&l==point)    {        sum[pos]=change;        return;    }    int mid=(l+r)>>1;    if(point<=mid)    update(point,change,l,mid,pos<<1);    else    update(point,change,mid+1,r,pos<<1|1);    pushup(pos);}int main(){    int T,cas=1;    scanf("%d",&T);    while(T--)    {        int n,m;        scanf("%d%lld",&n,&mod);        buildtree(1,n,1);        printf("Case #%d:\n",cas++);        for(int i=1;i<=n;i++)        {            int flag;            ll l;            scanf("%d%lld",&flag,&l);            if(flag==1)            update(i,l,1,n,1);            else            update(l,1,1,n,1);            printf("%lld\n",sum[1]);        }    }    return 0;}

 

hdu 5475 线段树


联系
我们