An easy problem
Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2957 Accepted Submission(s): 1095
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
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
2015 ACM/ICPC Asia Regional Shanghai Online
思路：
题目意思非常明确。本来是想着直接模拟求解。但是存在除法取余的操作，并且MOD并非是素数，所以逆元无法得解。因此排除这种思路。
我们利用线段树维护，每个叶子节点的值为当前的操作数，（若y == 1，就把叶子节点的值更新为操作数，若 y==2，就把叶子节点的值更新为1（相当于在之前操作的基础上做了除法））。对于每次询问，我们询问根节点的值即可。
#include <bits/stdc++.h>
using namespace std;
#define clr(a) memset(a,0,sizeof(a))
#define line cout<<"-----------------"<<endl;
typedef long long ll;
const int maxn = 1e5+10;
const int MAXN = 1e6+10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
const int N = 1010;
ll q,m,op[maxn],y;
struct node{
ll l,r,v;
ll mid (){
return (l + r) >> 1;
}
}tree[maxn<<2];
void pushup(ll x){
tree[x].v = tree[x<<1].v * tree[x<<1|1].v % m;
}
void build(ll x,ll l,ll r){
tree[x].l = l;
tree[x].r = r;
if(l == r){
tree[x].v = 1;
return ;
}
int mid = tree[x].mid();
build(x<<1,l,mid);
build(x<<1|1,mid+1,r);
pushup(x);
}
void update(ll x,ll pos,ll v){
if(tree[x].l == pos && tree[x].r == pos){
tree[x].v = v % m;
return ;
}
int mid = tree[x].mid();
if(pos <= mid) update(x<<1,pos,v);
else update(x<<1|1,pos,v);
pushup(x);
}
int main(){
int t;
scanf("%d",&t);
int T = 1;
while(t--){
printf("Case #%d:\n",T++);
scanf("%lld%lld",&q,&m);
build(1,1,q);
line;
for(int i=1.l;i<=20;i++){
printf("%lld ",tree[i].v);
}cout << endl;
line;
for(int i=1;i<=q;i++){
scanf("%lld%lld",&y,&op[i]);
if(y == 1) update(1,i,op[i]);
else update(1,op[i],1);
printf("%lld\n",tree[1].v);
line;
for(int i=1.l;i<=20;i++){
printf("%lld ",tree[i].v);
}cout << endl;
line;
}
}
return 0;
}