【线段树】单点更新 hdu 5475 An easy problem

An easy problem

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2322    Accepted Submission(s): 940

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

题意：其实X=1，输入n，m ；n行数据，m是模；每行数据输入两个操作，1，x和2，y；意义分别为乘以x、除以第y行数据所对应的x（题目保证第y行为1操作）输出每行数据计算后的结果X；

思路：因为题意没有确定m而且m不确定为素数，所以我们不能用逆元来求；利用线段树单点更新（将某个点更新为1），区间（1~n）求积；

代码：

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;typedef long long LL;const int N=100005;int n,M;struct NODE{    int l,r;    LL value;    int mid(){        return (l+r)>>1;    }}tree[N<<2];void BuildTree(int l,int r,int rt){    tree[rt].l=l;    tree[rt].r=r;    tree[rt].value=1;    if(l==r)    {        tree[rt].value=1;        return ;    }    int m=tree[rt].mid();    BuildTree(l,m,rt<<1);    BuildTree(m+1,r,rt<<1|1);    tree[rt].value=tree[rt<<1].value*tree[rt<<1|1].value%M;}void UpdateTree(int l,LL val,int rt){    if(tree[rt].l==l&&tree[rt].r==l)    {        tree[rt].value=val;        return ;    }    int m=tree[rt].mid();    if(l<=m) UpdateTree(l,val,rt<<1);    if(l>m) UpdateTree(l,val,rt<<1|1);    tree[rt].value=tree[rt<<1].value*tree[rt<<1|1].value%M;}LL Query(int l,int r,int rt){    if(tree[rt].l==l&&tree[rt].r==r)        return tree[rt].value;    int m=tree[rt].mid();    if(r<=m) return Query(l,r,rt<<1);    else if(l>m) return Query(l,r,rt<<1|1);    else return (Query(l,m,rt<<1)*Query(m+1,r,rt<<1|1))%M;}int main(){    int t;    scanf("%d",&t);    int cas=0;    while(t--)    {        scanf("%d%d",&n,&M);        BuildTree(1,n,1);        printf("Case #%d:\n",++cas);        for(int i=1;i<=n;i++)        {            int a;            LL b;            scanf("%d%I64d",&a,&b);            if(a==1)                UpdateTree(i,b,1);            else if(a==2)                UpdateTree(b,1,1);            printf("%I64d\n",tree[1].value);        }    }    return 0;}