题意：
给定n长的序列 m个操作
序列默认为 1, 2, 3···n
操作1：D [l,r] 把[l,r]区间增长 ：（ 1,2,3,4 进行 D [1,3]变成 1,1,2,2,3,3,4 ）
操作2：Q [l,r] 问区间[l,r] 上出现最多次数的数 的次数
线段树，维护每个区间的size 和叶子节点中最大的size
开始二分查找size的前缀和，逗了一场。。其实直接dfs就好了嘛 Orz
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
inline void rd(long long &n){
n = 0;
char c = getchar();
while(c < '0' || c > '9') c = getchar();
while(c >= '0' && c <= '9') n *= 10, n += (c - '0'),c = getchar();
}
inline void pt(long long n){
int len = 0,data[20];
while(n)
data[len++] = n%10, n /= 10;
if(!len) data[len++] = 0;
while(len--) putchar(data[len]+48);
}
#define N 51000
typedef long long ll;
inline int Mid(int x, int y){return (x+y)>>1;}
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define Lson(x) tree[x].l
#define Rson(x) tree[x].r
#define Siz(x) tree[x].siz
#define Val(x) tree[x].val
#define Lazy(x) tree[x].lazy
#define Max(x) tree[x].maxsize
struct node{
int l, r, val;
ll siz, lazy, maxsize;//siz是区间内的点数 maxsize是区间内最大点数 ,val是点的值
}tree[N<<2];
ll mul[60];
void push_down(int id){
if(Lazy(id)) {
Siz(L(id)) = Siz(L(id)) << Lazy(id);
Siz(R(id)) = Siz(R(id)) << Lazy(id);
Max(L(id)) *= mul[Lazy(id)];
Max(R(id)) *= mul[Lazy(id)];
Lazy(L(id)) += Lazy(id);
Lazy(R(id)) += Lazy(id);
Lazy(id) = 0;
}
}
void push_up(int id){
Max(id) = max(Max(L(id)), Max(R(id)));
Siz(id) = Siz(L(id)) + Siz(R(id));
}
void build(int l, int r, int id){
Lson(id) = l; Rson(id) = r; Lazy(id) = 0;
if(l == r){
Val(id) = l; Max(id) = Siz(id) = 1;
return ;
}
int mid = Mid(l, r);
build(l, mid, L(id)); build(mid+1, r, R(id));
push_up(id);
}
void updata_siz(int pos, ll he, int id){
push_down(id);
if(Lson(id) == Rson(id))
{
Siz(id) += he;
Max(id) += he;
return ;
}
int mid = Mid(Lson(id), Rson(id));
if(mid < pos)
updata_siz(pos, he, R(id));
else
updata_siz(pos, he, L(id));
push_up(id);
}
void updata_qujian(int l, int r, int id){
push_down(id);
if(l == Lson(id) && Rson(id) == r){
Lazy(id) ++;
Max(id) <<= 1LL;
Siz(id) <<= 1LL;
return ;
}
int mid = Mid(Lson(id), Rson(id));
if(mid < l)
updata_qujian(l, r, R(id));
else if(r <= mid)
updata_qujian(l, r, L(id));
else {
updata_qujian(l, mid, L(id));
updata_qujian(mid+1, r, R(id));
}
push_up(id);
}
ll query_siz(int l, int r, int id){
push_down(id);
if(l == Lson(id) && Rson(id) == r)
return Siz(id);
int mid = Mid(Lson(id), Rson(id));
if(mid < l)
return query_siz(l, r, R(id));
else if(r <= mid)
return query_siz(l, r, L(id));
else
return query_siz(l, mid, L(id)) + query_siz(mid+1, r, R(id));
}
ll query_max(int l, int r, int id){
push_down(id);
if(l == Lson(id) && Rson(id) == r)
return Max(id);
int mid = Mid(Lson(id), Rson(id));
if(mid < l)
return query_max(l, r, R(id));
else if(r <= mid)
return query_max(l, r, L(id));
else
return max(query_max(l, mid, L(id)), query_max(mid+1, r, R(id)));
}
int n, m;
int find(ll x, int id){
push_down(id);
if(Lson(id) == Rson(id))return Val(id);
if( x <= Siz(L(id)))
return find(x, L(id));
return find(x-Siz(L(id)), R(id));
}
char s[2]; ll l, r;
int main() {
mul[0] = 1; for(int i = 1; i < 60; i++)mul[i] = mul[i-1] * 2LL;
int T, Cas = 1; scanf("%d", &T);
while (T--) {
scanf("%d %d",&n,&m);
build(1, n, 1);
printf("Case #%d:\n", Cas++);
while(m--){
scanf("%s", s);
rd(l); rd(r);
int z = find(l, 1), y = find(r, 1);
if(z==y)
{
if(s[0] == 'D')
updata_siz(z, r-l+1, 1);
else
{pt(r-l+1);putchar('\n');}
continue;
}
ll Lsize = query_siz(1, z, 1) - l+1, Rsize = r - query_siz(1, y-1, 1);
if(s[0] == 'Q')
{
ll ans = max(Lsize, Rsize);
if(z + 1 <= y -1)
ans = max(ans, query_max(z+1, y-1, 1));
pt(ans); putchar('\n');
}
else
{
updata_siz(z, Lsize, 1);
updata_siz(y, Rsize, 1);
if(z + 1 <= y -1)
updata_qujian(z+1, y-1, 1);
}
}
}
return 0;
}
/*
99
10 99
D 1 4
Q 1 14
D 2 7
Q 1 20
D 3 8
Q 1 4
Q 1 12
Q 1 11
Q 10 17
D 1 3
D 1 3
Q 1 18
Q 1 33
D 1 33
D 1 33
D 1 99
D 1 99
Q 1 105
Q 1 200
Q 150 210
Q 75 113
D 1 10
Q 2 8
Q 2 50
D 90 200
Q 1 300
Q 2 50
Q 90 160
Q 45 65
Q 78 92
Q 100 200
ans:
2
4
4
8
7
5
10
10
105
160
50
39
7
49
71
21
15
101
*/
