hdoj5475An easy problem【线段树】

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1149    Accepted Submission(s): 563

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

 

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;const int maxn=100010;int MOD;long long sum[maxn<<2];void pushup(int rt){sum[rt]=(sum[rt<<1]*sum[rt<<1|1])%MOD;}void build(int l,int r,int rt){if(l==r){sum[rt]=1;return ;}int mid=(l+r)>>1;build(lson);build(rson);pushup(rt);}void update(int pos,long long k,int l,int r,int rt){if(l==r){sum[rt]=k;return ;}int mid=(l+r)>>1;if(pos<=mid)update(pos,k,lson);else update(pos,k,rson);pushup(rt);}long long query(int left,int right,int l,int r,int rt){if(l>right||r<left)return 1;if(left<=l&&r<=right){return sum[rt];}int mid=(l+r)>>1;return (query(left,right,lson)*query(left,right,rson))%MOD;}int main(){int t,i,j,k=1,n;scanf("%d",&t);while(t--){scanf("%d%d",&n,&MOD);long long a,b;build(1,n,1);printf("Case #%d:\n",k++);for(i=1;i<=n;++i){scanf("%lld%lld",&a,&b);if(a==1){update(i,b,1,n,1);printf("%lld\n",query(1,i,1,n,1));}else {update(b,1,1,n,1);printf("%lld\n",query(1,i,1,n,1));}}}return 0;}