HDU 5475（2015 ICPC上海站网络赛）--- An easy problem（线段树点修改）
题目链接
http://acm.hdu.edu.cn/showproblem.php?pid=5475
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.
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.
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.
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
Recommend
题意：输入Q和M，然后输入Q次操作，每次操作为一行数据op和s，定义一个数x=1。如果op=1表示x=x*s%mod 输出x，如果op=2 表示x除以第s次操作的s（第s次操作的op一定为1），输出x;
思路：线段树点修改，每次操作时，修改那个点，复杂度为log（n）,如果op=1，那么修改为s，如果op=2，那么第s次操作对应的点修改为1；
代码如下：
#include <iostream> #include <algorithm> #include <stdio.h> #include <cstring> #include <cmath> #include <map> #include <bitset> using namespace std; typedef long long LL; LL mod; LL tr[4*100005]; int p; LL add; void build(int l,int r,int i) { if(l==r) { tr[i]=1; return ; } int mid=(l+r)>>1; build(l,mid,i<<1); build(mid+1,r,i<<1|1); tr[i]=1; } void update(int l,int r,int i) { if(l==r) { tr[i]=add; return ; } int mid=(l+r)>>1; if(p<=mid) update(l,mid,i<<1); else update(mid+1,r,i<<1|1); tr[i]=(tr[i<<1]*tr[i<<1|1])%mod; } int main() { int T,Q,Case=1; cin>>T; while(T--) { scanf("%d%lld",&Q,&mod); printf("Case #%d:\n",Case++); build(1,Q,1); for(int i=1;i<=Q;i++) { int x; LL y; scanf("%d%lld",&x,&y); if(x==1) { p=i; add=y; } else { p=y; add=1; } update(1,Q,1); printf("%lld\n",tr[1]); } } return 0; }
