hdu

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($1 \leq T \leq 10$), 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. ($1 \leq Q \leq 10^5, 1 \leq M \leq 10^9$) 
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. ($0 < y \leq 10^9$) 
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

题意是每组测试给出Q,M，Q是查询数，每组查询给出op，n，有x初始值为1，若op=1则x乘上n，若op=2则x除以第n个查询中乘的数（这里对应的一定是op=1的n），每组查询输出X%M,

可用线段树，单点更新，区间查询，区间存放的是区间内元素的积

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define M 100005#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1ll q,mod;ll tree[M<<2];void pushup(ll rt){    tree[rt]=(tree[rt<<1]*tree[rt<<1|1])%mod;}void build(ll l,ll r,ll rt){    if(l==r)    {        tree[rt]=1;        return;    }    ll m=(l+r)>>1;    build(lson);    build(rson);    tree[rt]=1;}void update(ll p,ll v,ll l,ll r,ll rt){    if(l==r)    {        tree[rt]=v;        return;    }    ll m=(l+r)>>1;    if(p<=m)        update(p,v,lson);    else        update(p,v,rson);    pushup(rt);}int main(){    int T,cas=0;    ll op,n,i,v,p;    scanf("%d",&T);    while(T--)    {        scanf("%lld%lld",&q,&mod);        build(1,q,1);        //memset(tree,1,sizeof(tree)); //longlong不能用memset?        printf("Case #%d:\n",++cas);        for(i=1;i<=q;i++)        {            scanf("%lld%lld",&op,&n);            if(op==1)            {               p=i;               v=n;            }else            {                p=n;                v=1;            }            update(p,v,1,q,1);            printf("%lld\n",tree[1]);        }    }    return 0;}

还可以直接模拟，注意因为题目中不断对x取余所以进行除法时已经我发得到正确的值，所以遇到除法从头开始乘就行。

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define M 100005ll q,m;ll number[M];bool vis[M];int main(){    int T,cas=0;    ll op,n,i,j;    scanf("%d",&T);    while(T--)    {        scanf("%lld%lld",&q,&m);        printf("Case #%d:\n",++cas);        ll x=1;        memset(vis,0,sizeof(vis));        for(i=1;i<=q;i++)        {            scanf("%lld%lld",&op,&n);            number[i]=n;            if(op==1)            {                vis[i]=true;                x=(x*n)%m;  //计算%是很费时的，即如果这里换为x=(x%m*n%m)%m，会处在超时边缘            }else            {                x=1;                for(j=1;j<i;j++)                {                    if(j==n)                        vis[j]=false;                    else                    {                        if(vis[j]==true)                            x=(x*number[j])%m;                    }                }            }            printf("%lld\n",x);        }    }    return 0;}