hdu5475__An easy problem(线段树单点更新)

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 593    Accepted Submission(s): 342

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

 

题意：输入T代表T组样例，每组输入一个q，m分别代表有q个操作，每次操作后结果要对m取余。每次操作输入两个数a，b，若a=1，则代表结果是要乘以b，若a=2，则代表结果要除以第b次操作输入的'b'。

想法：线段树，把q次操作当成是q个节点，第i次操作时，当a=1时，把i节点对应的值更新成b，当a=2时，把b节点对应的值更新成1。

代码如下：

#include<cstdio>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define maxn 111111typedef long long ll;int T,q,M;ll col[maxn<<2];void pushup(int rt){    col[rt]=col[rt<<1]*col[rt<<1|1]%M;}void build(int l,int r,int rt){    if(l==r)    {        col[rt]=1;        return ;    }    int m=(l+r)>>1;    build(lson);    build(rson);    pushup(rt);}void update(int p,int c,int l,int r,int rt){    if(l==r&&l==p)    {        col[rt]=c;        return ;    }    int m=(l+r)>>1;    if(p<=m) update(p,c,lson);    else update(p,c,rson);    pushup(rt);}int main(){    int a,b,cas=1;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&q,&M);        build(1,q,1);        printf("Case #%d:\n",cas++);        for(int i=1; i<=q; i++)        {            scanf("%d%d",&a,&b);            if(a==1)            {                update(i,b,1,q,1);            }            else if(a==2)            {                update(b,1,1,q,1);            }            printf("%d\n",col[1]);        }    }    return 0;}