题目如下:
给定一个序列,有多次询问,每次查询区间里小于等于某个数的元素的个数
即对于询问 (l,r,x),你需要输出 的值
其中 [exp] 是一个函数,它返回 1 当且仅当 exp 成立,其中 exp 表示某个表达式
这个题可以使用两种方法来解决:
树状数组:由于每一个数都有其地址的编号,所以先记录没个数的编号,然后对这组数进行排序,对于(l, r,x)进行对多次输入也分别记录他们的地址,这样就可以分别输出符合条件的数,然后就是树状数组的基本更新和求和操作。每一次比较的是比较val,但是更新的是他的编号,这样就可以直接查询不怕出现错误;
代码如下:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#define pi acos(-1)
#define e exp(1)
#define For(i, a, b) for(int (i) = (a); (i) <= (b); (i) ++)
#define Bor(i, a, b) for(int (i) = (b); (i) >= (a); (i) --)
#define max(a,b) (((a)>(b))?(a):(b))
#define min(a,b) (((a)<(b))?(a):(b))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define eps 1e-7
#define INF 0x3f3f3f3f
#define inf -2100000000
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxn = 1e5 + 10;
const double EPS = 1e-10;
const ll p = 1e7+9;
const ll mod = 1e9+7;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
inline int read(){
int ret=0,f=0;char ch=getchar();
while(ch>'9'||ch<'0') f^=ch=='-',ch=getchar();
while(ch<='9'&&ch>='0') ret=ret*10+ch-'0',ch=getchar();
return f?-ret:ret;
}
int n, m;
struct node{
int l, r, val;
int id;
}a[maxn],b[maxn];
int sum[maxn];
int ans[maxn];
int lowbit(int rt){
return rt & (-rt);
}
void add(int rt){
while(rt <= n){
sum[rt] += 1;
rt += lowbit(rt);
}
return ;
}
int Sum(int rt){
int ans = 0;
while(rt > 0){
ans += sum[rt];
rt -= lowbit(rt);
}
return ans;
}
bool cmp(node x, node y){
return x.val < y.val;
}
int main(){
ios::sync_with_stdio(false);
cin >> n >> m;
for(int i = 1; i <= n; i++){
cin >> a[i].val;
a[i].id = i;
}
sort(a+1, a+n+1, cmp);
for(int i = 1; i <= m; i++){
cin >> b[i].l >> b[i].r >> b[i].val;
b[i].id = i;
}
sort(b+1, b+1+m, cmp);
int q = 1;
for(int i = 1; i <= m; i++){
while(a[q].val <= b[i].val && q <= n ){
add(a[q].id);
q++;
}
ans[b[i].id] = Sum(b[i].r) - Sum(b[i].l-1);
}
for(int i = 1; i <= m; i++){
cout << ans[i] << endl;
}
return 0;
}
线段树的方法:
线段树和树状数组的操作是一样是的,都是值比较大小然后把他的地址更新:然后使用区间求和得到值:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#define pi acos(-1)
#define e exp(1)
#define For(i, a, b) for(int (i) = (a); (i) <= (b); (i) ++)
#define Bor(i, a, b) for(int (i) = (b); (i) >= (a); (i) --)
#define max(a,b) (((a)>(b))?(a):(b))
#define min(a,b) (((a)<(b))?(a):(b))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define eps 1e-7
#define INF 0x3f3f3f3f
#define inf -2100000000
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxn = 1e5 + 10;
const double EPS = 1e-10;
const ll p = 1e7+9;
const ll mod = 1e9+7;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
inline int read(){
int ret=0,f=0;char ch=getchar();
while(ch>'9'||ch<'0') f^=ch=='-',ch=getchar();
while(ch<='9'&&ch>='0') ret=ret*10+ch-'0',ch=getchar();
return f?-ret:ret;
}
struct node{
int l,r,val,id;
}b[maxn],a[maxn];
int ans[maxn],tree[maxn << 2];
bool cmp(node a, node b){
return a.val < b.val;
}
void push_up(int rt){
tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}
void update(int p, int l, int r, int rt){
if(l == r){
tree[rt]++;
return ;
}
int mid = (l+r) >> 1;
if(p <= mid) update(p, lson);
else update(p, rson);
push_up(rt);
}
int query(int a,int b,int l,int r,int rt){
if(a<=l && r<=b){
return tree[rt];
}
int mid = (l+r) >> 1;
int ans = 0;
if(a <= mid) ans += query(a, b, lson);
if(b > mid)ans += query(a, b, rson);
return ans;
}
int main(){
int n,m;
cin >> n >> m;
for(int i = 1; i <= n; i++){
cin >> a[i].val;
a[i].id = i;
}
sort(a + 1, a + n + 1, cmp);
for(int i = 1; i <= m; i++){
cin >> b[i].l >> b[i].r >> b[i].val;
b[i].id = i;
}
sort(b + 1,b + m + 1, cmp);
int q = 1;
for(int i = 1; i <= m; i++){
while(q <= n && a[q].val <= b[i].val){
update(a[q].id, 1, n, 1);
//cout << "+++" << a[q].id << " " << tree[a[q].id] << endl;
q++;
}
ans[b[i].id] = query(b[i].l, b[i].r, 1, n, 1);
}
for(int i=1; i <= m; i++){
cout<<ans[i]<<endl;
}
}
脑子要变通,不能太死板: