2015上海网络赛 HDU 5475 An easy problem 线段树

题意就不说了

思路：线段树，维护区间乘积。2操作就将要除的点更新为1.

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #include<vector>
 #include<set>
 #include<string>
 #define inf 0x3f3f3f3f
 #define LL long long
 #define mid(a,b) ((a+b)>>1)
 #define ll long long
 #define maxn 110000
 #define IN freopen("in.txt","r",stdin);
 using namespace std;

 LL mod;
 LL num[maxn];
 struct Tree
 {
     int left,right;
     LL pro;    /*pro为区间和，可改为最大值最小值等*/
 }tree[maxn<<]; /*四倍大小*/

 /*递归建树*/
 void build(int i,int left,int right)
 {
     tree[i].left=left;
     tree[i].right=right;

     if(left==right){
         tree[i].pro=num[left]%mod;
         return ;
     }

     int mid=mid(left,right);

     build(i<<,left,mid);
     build(i<<|,mid+,right);

     tree[i].pro=(tree[i<<].pro*tree[i<<|].pro)%mod;
 }

 /*单点修改，d为改变量，两函数可共存*/
 void update(int i,int x,LL d)
 {
     if(tree[i].left==tree[i].right){
         tree[i].pro=d;
         return;
     }

     int mid=mid(tree[i].left,tree[i].right);

     if(x<=mid) update(i<<,x,d);
     else update(i<<|,x,d);

     tree[i].pro=(tree[i<<].pro*tree[i<<|].pro)%mod;
 }

 /*区间结果查询*/
 LL query(int i,int left,int right)
 {
     if(tree[i].left==left&&tree[i].right==right)
         return tree[i].pro%mod;

     int mid=mid(tree[i].left,tree[i].right);

     if(right<=mid) return query(i<<,left,right);
     else if(left>mid) return query(i<<|,left,right);
     else return (query(i<<,left,mid)*query(i<<|,mid+,right))%mod;
 }

 int main(int argc, char const *argv[])
 {
     //IN;

     int t,ca=;scanf("%d",&t);
     while(t--)
     {
         printf("Case #%d:\n",ca++);
         int n;scanf("%d %lld",&n,&mod);
         //fill(num,num+maxn,1);
         int type[maxn];LL val[maxn];
         for(int i=;i<=n;i++){
             scanf("%d %lld",&type[i],&val[i]);
             if(type[i]==) num[i]=val[i];
             if(type[i]==) num[i]=;
         }
         build(,,n);
         for(int i=;i<=n;i++){
             if(type[i]==){
                 LL ans=query(,,i);
                 printf("%lld\n",ans%mod);
             }
             else if(type[i]==){
                 update(,val[i],1LL);
                 LL ans=query(,,i);
                 printf("%lld\n",ans%mod);
             }
         }

     }

     return ;
 }