dong2022214184的博客

1.普通二分（无重复值）[L,R]#include <iostream>using namespace std;int arr[150],n;int erfen(int target){ int l=0,mid; int r=n-1; while(l<=r) { mid=(l+r)/2; if(arr[mid]==target) { return mid;

1.取第一个while (left < right) { mid = left + (right - left) / 2; if (arr[mid] < target) left = mid + 1; else right = mid; } if (arr[left] == target) return left; else return -1;注意点（1）.取第一个时由于程序本身向下取整，且当相同时r=m

int age;int grade;bool operator <(const stu &b)const{ if(grade==b.grade) { string l1=name,l2=b.name; int tmp= l1.compare(l2); return tmp<0; } return grade>b.grade;}}s[1000];

归并排序及求逆序对数#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <queue>#define mem(x,y) memset(x,y,sizeof x)#define fo(i,s,t) for(int ...

K. Center枚举两个点的中点作为轴对称的点就可以了#include <algorithm>#include <string.h>#include <stdio.h>#include <map>#include <utility>#define ll long long#define PII pair<ll,ll...

l为半周长，s为面积double l = (a + b + c) / 2;double s = sqrt(l*(l - a)(l - b)(l - c));