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.
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
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: ",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 ",sum[1]); } } return 0; }