2015
02-21

# Jewel

Jimmy wants to make a special necklace for his girlfriend. He bought many beads with various sizes, and no two beads are with the same size. Jimmy can’t remember all the details about the beads, for the necklace is so long. So he turns to you for help.

Initially, there is no bead at all, that is, there is an empty chain. Jimmy always sticks the new bead to the right of the chain, to make the chain longer and longer. We number the leftmost bead as Position 1, and the bead to its right as Position 2, and so on. Jimmy usually asks questions about the beads’ positions, size ranks and actual sizes. Specifically speaking, there are 4 kinds of operations you should process:

Insert x
Put a bead with size x to the right of the chain (0 < x < 231, and x is different from all the sizes of beads currently in the chain)
Query_1 s t k
Query the k-th smallest bead between position s and t, inclusive. You can assume 1 <= s <= t <= L, (L is the length of the current chain), and 1 <= k <= min (100, t-s+1)
Query_2 x
Query the rank of the bead with size x, if we sort all the current beads by ascent order of sizes. The result should between 1 and L (L is the length of the current chain)
Query_3 k
Query the size of the k-th smallest bead currently (1 <= k <= L, L is the length of the current chain)

There are several test cases in the input. The first line for each test case is an integer N, indicating the number of operations. Then N lines follow, each line contains one operation, as described above.

You can assume the amount of "Insert" operation is no more than 100000, and the amounts of "Query_1", "Query_2" and "Query_3" are all less than 35000.
There are several test cases in the input. The first line for each test case is an integer N, indicating the number of operations. Then N lines follow, each line contains one operation, as described above.

You can assume the amount of "Insert" operation is no more than 100000, and the amounts of "Query_1", "Query_2" and "Query_3" are all less than 35000.Query the rank of the bead with size x, if we sort all the current beads by ascent order of sizes. The result should between 1 and L (L is the length of the current chain)
Query_3 k
Query the size of the k-th smallest bead currently (1 <= k <= L, L is the length of the current chain)

There are several test cases in the input. The first line for each test case is an integer N, indicating the number of operations. Then N lines follow, each line contains one operation, as described above.

You can assume the amount of "Insert" operation is no more than 100000, and the amounts of "Query_1", "Query_2" and "Query_3" are all less than 35000.
There are several test cases in the input. The first line for each test case is an integer N, indicating the number of operations. Then N lines follow, each line contains one operation, as described above.

You can assume the amount of "Insert" operation is no more than 100000, and the amounts of "Query_1", "Query_2" and "Query_3" are all less than 35000.Query the rank of the bead with size x, if we sort all the current beads by ascent order of sizes. The result should between 1 and L (L is the length of the current chain)
Query_3 k
Query the size of the k-th smallest bead currently (1 <= k <= L, L is the length of the current chain)

10
Insert 1
Insert 4
Insert 2
Insert 5
Insert 6
Query_1 1 5 5
Query_1 2 3 2
Query_2 4
Query_3 3
Query_3 1

Case 1:
10
3
5

HintThe answers for the 5 queries are 6, 4, 3, 4, 1, respectively. 

# 题目大意

1、Insert x 在序列尾部插入x

2、Query_1 s t k 查询区间[s,t]的第k小

3、Query_2 x 查询x的在序列中排名

4、Query_3 k 查询序列中的第k小

# 代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
#define MAXN 100005
#define lson l,m,ls[s]
#define rson m+1,r,rs[s]
typedef long long LL;
int ls[20*MAXN],rs[20*MAXN];
int cnt[20*MAXN],T[MAXN];
int a[MAXN],num[MAXN];
char op[2*MAXN][15];
int L[2*MAXN],R[2*MAXN],K[2*MAXN];
int tot;
vector<int>ivec;
void build(int l,int r,int &s)
{
s=++tot;
cnt[s]=0;
if(l==r) return;
int m=(l+r)>>1;
build(lson);
build(rson);
}
void update(int last,int p,int l,int r,int &s)
{
s=++tot;
ls[s]=ls[last],rs[s]=rs[last];
cnt[s]=cnt[last]+1;
if(l==r) return;
int m=(l+r)>>1;
if(p<=m) update(ls[last],p,lson);
else     update(rs[last],p,rson);
}
int query(int ss,int tt,int l,int r,int k)
{
if(l==r) return r;
int sum=cnt[ls[tt]]-cnt[ls[ss]];
int m=(l+r)>>1;
if(sum>=k) return query(ls[ss],ls[tt],l,m,k);
else       return query(rs[ss],rs[tt],m+1,r,k-sum);
}
int main()
{
int m,kase=0;
while(scanf("%d",&m)!=EOF)
{
memset(cnt,0,sizeof(cnt));
int len=0,n;
LL qy1=0,qy2=0,qy3=0;
ivec.clear();
for(int i=0; i<m; i++)
{
scanf("%s%d",op[i],&L[i]);
if(!strcmp(op[i],"Insert"))
{
++len;
num[len]=a[len]=L[i];
}
else if(!strcmp(op[i],"Query_1"))
scanf("%d%d",&R[i],&K[i]);
}
sort(a+1,a+len+1);
n=len;
len=unique(a+1,a+len+1)-a-1;
for(int i=1; i<=n; i++) num[i]=lower_bound(a+1,a+len+1,num[i])-a;
tot=0;
build(1,len,T[0]);
for(int i=1; i<=n; i++) update(T[i-1],num[i],1,len,T[i]);
for(int i=0; i<m; i++)
{
if(!strcmp(op[i],"Insert"))
ivec.insert(lower_bound(ivec.begin(),ivec.end(),L[i]),L[i]);
else if(!strcmp(op[i],"Query_1"))
qy1+=a[query(T[L[i]-1],T[R[i]],1,len,K[i])];
else if(!strcmp(op[i],"Query_2"))
qy2+=lower_bound(ivec.begin(),ivec.end(),L[i])-ivec.begin()+1;
else
qy3+=ivec[L[i]-1];
}
printf("Case %d:\n",++kase);
printf("%I64d\n%I64d\n%I64d\n",qy1,qy2,qy3);
}
return 0;
}

1. 为什么for循环找到的i一定是素数叻，而且约数定理说的是n=p1^a1*p2^a2*p3^a3*…*pk^ak，而你每次取余都用的是原来的m，也就是n