Given a integers x = 1, you have to apply Q (Q ≤ 100000) operations: Multiply, Divide.
Input
First line of the input file contains an integer T(0 < T ≤ 10) that indicates how many cases of inputs are there.
The description of each case is given below:
The first line contains two integers Q and M. The next Q lines contains the operations in ith line following form:
M yi: x = x * yi.
N di: x = x / ydi.
It’s ensure that di is different. That means you can divide yi only once after yi came up.
0 < yi ≤ 10^9, M ≤ 10^9
Output
For each operation, print an integer (one per line) x % M.
Sample Input
1
10 1000000000
M 2
D 1
M 2
M 10
D 3
D 4
M 6
M 7
M 12
D 7
Sample Output
2
1
2
20
10
1
6
42
504
84
【分析】
针对一个数组的操作，即对一个区间。可以用线段树去进行维护。初始化建树，叶子节点的值为1，维护每段区间上各个元素的乘积sum。M yi，将第i个元素的值改为yi。N di，将第di个元素的值改为1。输出即查询区间[1,Q]的sum值。也就是变成了单点更新、区间查询问题。
时间复杂度为O（QlongQ）。
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<stack>
#include<sstream>
#include<list>
#include<bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e5;
const double eps = 1e-8;
LL mod,val;
LL sum[maxn*4];
void update(int p,int val,int l,int r,int rt)
{
if(l==r)
{
sum[rt]=val;
return ;
}
int m=(l+r)/2;
if(p<=m)
update(p,val,l,m,rt*2);
else
update(p,val,m+1,r,rt*2+1);
sum[rt] = sum[rt*2] * sum[rt*2+1] % mod;
}
//char op[10];
int main()
{
//ios::sync_with_stdio(false);
int t,q;
scanf("%d",&t);
while(t--)
{
scanf("%d%lld",&q,&mod);
for(int i=1;i<=4*maxn;i++) sum[i]=1;
for(int i=1;i<=q;i++)
{
int x;char op[10];
scanf("%s%d",op,&x);
if(op[0]=='M')
{
update(i,x,1,maxn,1);
printf("%lld
",sum[1]);
}
else
{
update(x,1,1,maxn,1);
printf("%lld
",sum[1]);
}
}
}
return 0;
}
/*
1
10 1000000000
M 2
D 1
M 2
M 10
D 3
D 4
M 6
M 7
M 12
D 7
2
1
2
20
10
1
6
42
504
84
*/