3月22日-课堂笔记
- 非降序序列二分查找等于 x x x 的数下标
int find(int x, int l, int r) {
while(l < r) {
int mid = (l + r) / 2;
if(x <= a[mid]) r = mid;
else l = mid + 1;
}
return l;
}
- 非降序可重序列下标最小 ≥ x \geq x ≥x 的元素
int find(int x, int l, int r) {
while(l < r) {
int mid = (l + r) / 2;
if(a[mid] >= x) r = mid;
else l = mid + 1;
}
return l;
}
- C++ STL 的使用
vector<int> a(n);
int id = lower_bound(a.begin(), a.end(), x) - a.begin();
int id = upper_bound(a.begin(), a.end(), x) - a.begin();
int a[N];
int id = lower_bound(a + 1, a + n + 1, x) - a;
int id = upper_bound(a + 1, a + n + 1, x) - a;
- a中元素 x x x 出现的次数 O ( log n ) \mathcal{O}(\log n) O(logn)
int count = upper_bound(a.begin(),a.end(),x) – lower_bound(a.begin(),a.end(),x);
练习题解
P1102 A-B 数对
题目链接：P1102
参考思路
对于给定的 A − B = C A-B=C A−B=C，其中 C C C已知，那么我们可以枚举 A A A二分查找 B B B的范围，其中
C++参考代码
//用系统库函数
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int a[N];
signed main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int n, C;
cin >> n >> C;
for(int i = 1; i <= n; ++ i) cin >> a[i];
sort(a + 1, a + n + 1);
long long ans = 0;
for(int i = 1; i <= n; ++ i) {
int L = lower_bound(a + 1, a + n + 1, a[i] - C) - a;
int R = upper_bound(a + 1, a + n + 1, a[i] - C) - a - 1;
ans += R - L + 1;
}
cout << ans << endl;
return 0;
}
//方便大家理解手动实现了一下函数
#include <bits/stdc++.h>
using namespace std;
int lowerBound(const vector<int>& a, int key) {
int low = 0, high = a.size();
while (low < high) {
int mid = (low + high) / 2;
if (a[mid] < key) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
int upperBound(const vector<int>& a, int key) {
int low = 0, high = a.size();
while (low < high) {
int mid = (low + high) / 2;
if (a[mid] <= key) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
int main() {
int n, C;
cin >> n >> C;
vector<int> a(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
sort(a.begin(), a.end());
long long ans = 0;
for (int i = 0; i < n; ++i) {
int L = lowerBound(a, a[i] - C);
int R = upperBound(a, a[i] - C);
ans += R - L;
}
cout << ans << endl;
return 0;
}
Java参考代码
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n, C;
n = scanner.nextInt();
C = scanner.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; ++i) {
a[i] = scanner.nextInt();
}
Arrays.sort(a);
long ans = 0;
for (int i = 0; i < n; ++i) {
int L = lowerBound(a, a[i] - C);
int R = upperBound(a, a[i] - C);
ans += R - L;
}
System.out.println(ans);
scanner.close();
}
static int lowerBound(int[] a, int key) {
int low = 0;
int high = a.length;
while (low < high) {
int mid = (low + high) / 2;
if (a[mid] < key) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
static int upperBound(int[] a, int key) {
int low = 0;
int high = a.length;
while (low < high) {
int mid = (low + high) / 2;
if (a[mid] <= key) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
}
Python参考代码
def lower_bound(a, key):
low, high = 0, len(a)
while low < high:
mid = (low + high) // 2
if a[mid] < key:
low = mid + 1
else:
high = mid
return low
def upper_bound(a, key):
low, high = 0, len(a)
while low < high:
mid = (low + high) // 2
if a[mid] <= key:
low = mid + 1
else:
high = mid
return low
def main():
n, C = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
ans = 0
for i in range(n):
L = lower_bound(a, a[i] - C)
R = upper_bound(a, a[i] - C)
ans += R - L
print(ans)
if __name__ == "__main__":
main()
P1678 烦恼的高考志愿
题目链接：P1678
参考思路
- 题目意思其实就是求某一个学生和某一个分数线最小的差的绝对值，求和。
- 那么我们先对学校分数线 a a a数组排序，然后对于每一个学生 b i b_i bi去学校分数也就是 a a a数组中二分查找学生分数 b i b_i bi左右两边最近的数即：最后一个小于等于 b i b_i bi的数和第一个大于等于 b i b_i bi的数。
- 特别注意特殊处理当所有的分数都高于或者低于学生分数的情况。
C++参考代码
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N], b[N];
int findr(int t, int l, int r) {
while (l < r) {
int mid = (l + r) / 2;
if (t <= a[mid]) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
int findl(int t,int l, int r) {
while (l < r) {
int mid = (l + r + 1) / 2;
if (t >= a[mid]) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}
int main() {
int m, n;
cin >> m >> n;
for (int i = 0; i < m; i++)
cin >> a[i];
for (int i = 0; i < n; i++)
cin >> b[i];
sort(a, a + m);
long long res = 0;
for (int i = 0; i < n; i++) {
if (b[i] <= a[0])
res += abs(b[i] - a[0]);
else if (b[i] >= a[m - 1])
res += abs(b[i] - a[m - 1]);
else {
int l = a[findr(b[i], 0, m)];
int r = a[findl(b[i], 0, m)];
int ls = abs(l - b[i]);
int rs = abs(r - b[i]);
if (ls < rs) {
res += ls;
} else {
res += rs;
}
}
}
cout << res << endl;
return 0;
}
Java参考代码
import java.util.*;
public class Main {
static int[] a = new int[100010];
static int[] b = new int[100010];
static int findr(int t, int l, int r) {
while (l < r) {
int mid = (l + r) / 2;
if (t <= a[mid]) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
static int findl(int t, int l, int r) {
while (l < r) {
int mid = (l + r + 1) / 2;
if (t >= a[mid]) {
l = mid;
} else {
r = mid - 1;
}
}
return l;
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int m = scanner.nextInt();
int n = scanner.nextInt();
for (int i = 0; i < m; i++)
a[i] = scanner.nextInt();
for (int i = 0; i < n; i++)
b[i] = scanner.nextInt();
Arrays.sort(a, 0, m);
long res = 0;
for (int i = 0; i < n; i++) {
if (b[i] <= a[0])
res += Math.abs(b[i] - a[0]);
else if (b[i] >= a[m - 1])
res += Math.abs(b[i] - a[m - 1]);
else {
int l = a[findr(b[i], 0, m)];
int r = a[findl(b[i], 0, m - 1)];
int ls = Math.abs(l - b[i]);
int rs = Math.abs(r - b[i]);
if (ls < rs) {
res += ls;
} else {
res += rs;
}
}
}
System.out.println(res);
}
}
Python参考代码
def findr(t, l, r):
while l < r:
mid = (l + r) // 2
if t <= a[mid]:
r = mid
else:
l = mid + 1
return l
def findl(t, l, r):
while l < r:
mid = (l + r + 1) // 2
if t >= a[mid]:
l = mid
else:
r = mid - 1
return l
m, n = map(int, input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
a.sort()
res = 0
for i in range(n):
if b[i] <= a[0]:
res += abs(b[i] - a[0])
elif b[i] >= a[m - 1]:
res += abs(b[i] - a[m - 1])
else:
l = a[findr(b[i], 0, m)]
r = a[findl(b[i], 0, m - 1)]
ls = abs(l - b[i])
rs = abs(r - b[i])
if ls < rs:
res += ls
else:
res += rs
print(res)
