BZOJ 1012 最大数maxnumber 线段树

题目链接：

https://www.lydsy.com/JudgeOnline/problem.php?id=1012

题目大意：

见链接

思路：

直接用线段树模拟一下就可以了。

 #include<bits/stdc++.h>
 #define IOS ios::sync_with_stdio(false);//不可再使用scanf printf
 #define Max(a, b) ((a) > (b) ? (a) : (b))//禁用于函数，会超时
 #define Min(a, b) ((a) < (b) ? (a) : (b))
 #define Mem(a) memset(a, 0, sizeof(a))
 #define Dis(x, y, x1, y1) ((x - x1) * (x - x1) + (y - y1) * (y - y1))
 #define MID(l, r) ((l) + ((r) - (l)) / 2)
 #define lson ((o)<<1)
 #define rson ((o)<<1|1)
 #pragma comment(linker, "/STACK:102400000,102400000")//栈外挂
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }

 typedef long long ll;
 const int maxn =  + ;
 const int maxm =  + ;
 const int mod = ;//const引用更快，宏定义也更快
 const int INF = ;
 struct node
 {
     ll l, r, x;
 }tree[maxn];
 void build(int o, ll l, ll r)
 {
     tree[o].l = l, tree[o].r = r;
     if(l == r)return;
     ll m = MID(l, r);
     build(lson, l, m);
     build(rson, m + , r);
 }
 ll p, v;
 void update(int o)
 {
     if(tree[o].l == tree[o].r)
     {
         tree[o].x = v;
         return;
     }
     if(p <= tree[lson].r)update(lson);
     else update(rson);
     tree[o].x = Max(tree[lson].x, tree[rson].x);
 }
 ll ql, qr;
 ll ans;
 void query(int o)
 {
     if(ql <= tree[o].l && qr >= tree[o].r)
     {
         ans = Max(ans, tree[o].x);
         return;
     }
     if(ql <= tree[lson].r)query(lson);
     if(qr >= tree[rson].l)query(rson);
 }
 int main()
 {
     ll n, d;
     scanf("%lld%lld", &n, &d);
     build(, , n);
     p = ;
     ll t = , x;
     char s[];
     while(n--)
     {
         scanf("%s%lld", s, &x);
         if(s[] == 'A')
         {
             v = t + x;
             v %= d;
             update();
             p++;
         }
         else
         {
             ql = p - x;
             qr = p - ;
             ans = ;
             query();
             t = ans;
             printf("%lld\n", ans);
         }
     }
     return ;
 }