hdu 5475(2015上海网赛) An easy problem

题目;http://acm.hdu.edu.cn/showproblem.php?pid=5475

题意就是给X赋初值1，然后给Q个操作，每个操作对应一个整数M；如果操作是1则将X乘以对应的M，如果是2则除以第M次操作对应的M'，求最后X的值对给定值取摸的结果,

直接暴力会爆long long,用数组存每一步的话 会超时,所以用线段树优化进行区间更新

 #include<cstdio>
 using namespace std;
 typedef long long ll;
 struct point {
     int l,r;
     ll sum;
 };
 point tree[*];
 int n,m;
 void build(int i,int left,int right)
 {
     tree[i].l=left;
     tree[i].r=right;
     if (left==right)
     {
         tree[i].sum=;
         return;
     }
     int mid=(left+right)>>;
     build(i<<,left,mid);
     build(i<<|,mid+,right);
     tree[i].sum=(tree[i<<].sum*tree[i<<|].sum)%m;
 }
 void update(int i,int num,int val)
 {
     if (tree[i].l==tree[i].r)
     {
         tree[i].sum=val;
         return;
     }
     int mid=(tree[i].l+tree[i].r)>>;
     if (num<=mid) update(i<<,num,val);
     else update(i<<|,num,val);
     tree[i].sum=(tree[i<<].sum*tree[i<<|].sum)%m;
 }
 ll _find(int i,int left,int right)
 {
     ll ans=;
     if (left<=tree[i].l&&tree[i].r<=right)
         return tree[i].sum;
     int mid=(tree[i].l+tree[i].r)>>;
     if (left<=mid) ans*=_find(i<<,left,right)%m;
     if (right>mid) ans*=_find(i<<|,left,right)%m;
     return ans%m;
 }
 int main()
 {
     int t,x,y,i,w=;
     scanf("%d",&t);
     while (t--)
     {
         printf("Case #%d:\n",w++);
         scanf("%d %d",&n,&m);
         build(,,n);
         for (i=;i<=n;i++)
         {
             scanf("%d %d",&x,&y);
             if (x==)
             {
                 update(,i,y);
                 printf("%I64d\n",_find(,,i)%m);
             }
             else
             {
                 update(,y,);
                 printf("%I64d\n",_find(,,i)%m);
             }
         }
     }
     return ;
 }