_计算机好苦

当ifstream File(filename);只要不关闭(file.close)这个文件都会继续往下读多个文件同时读取ifstream Files[9];Files[i].open(FileName, ios::in);

记录一下 ，每次都忘很烦C++数据类似这样的头文件:#pragma once#include <fstream>#include <iostream>#include <vector>#include <string>using namespace std;double Vertx[20], Verty[20];ifstream MyFile("E:\\大二下\\上机代码\\算法课设上机\\SatCoverInfo_0.txt");v

```cpp#include <stdio.h>#include <string.h> #include <stdlib.h> typedef int Position;#define NotFound -1 void BuildMatch( char *pattern, int *match ){ Position i, j; int m = strlen(pattern); match[0] = -1; f

求a0+a1x+a2x^ 2+…anx^ n数组a存系数int QinJiuShao(int n,double a[], int x){ int i; double p = a[n]; for (i = n; i > 0; i--) { p = a[i-1] + x * p; } return p;}

#include<iostream>#include<queue> //试着加加constusing namespace std;const int defaultSize=10;template<class T>class MinHeap{public: MinHeap(int sz = defaultSize); MinHeap(T arr[], int n); ~MinHeap() { delete[] heap; } bool push(co

#include<iostream>using namespace std;int a[1000000];int max3(int a, int b, int c){ if (a > b) { if (a > c) return a; else return c; } else { if (b > c) return b; else return c; }}int fenzhi(int a[], int left,

#include<iostream>using namespace std;#include<vector>struct node{ int data; node* next;};typedef node* list;list init(){ list a = new node; a->next = NULL; return a;}list insert(list a,int i,int x){ list p = a; for (int k

#include<iostream>using namespace std;#include<vector>struct node{ int data; node* next;};typedef node* list;list reverse(list a){ list pre = NULL, cur = a->next; list temp; while (cur != NULL) { temp = cur->next; cur-&g

总共 n个人 从s号开始，m个删去一个#include<iostream>#include<vector>using namespace std;int main(){ vector<int>v; int n, s, m; cin >> n >> s >> m; for (int i = 1; i <= n; i++) { v.push_back(i); } auto it = v.begin() +

这是stl deque的测试 仅实现部分功能这是利用首位相接的数组 [循环队列]仿stl双端队列deque将得到一样的结果#include<iostream>#include<unordered_map>using namespace std;template<class T>class Deque{ int start, rear; T* elements; int maxsize ;public: Deque(); ~Deque()

#include<iostream>using namespace std;int m, n; //终点坐标struct node{ int x, y;};int map[5][3];int mark[5][3];node Node[8] = { {-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};bool seekPath(int x, int y){ int xx, yy; if (x == m &.

#include<iostream>#include<stack>using namespace std;template<class T>class Stack{ T* elements; int topPtr; int maxsize;public: Stack(int size = 50); ~Stack(){ delete[]elements; } void push(const T&x); void pop(); T top(

递归的适用范围代替多重循环用递归解决递归形式的问题用递归将问题分解为规模更小的子问题汉诺塔问题#include<iostream>using namespace std;void digui(int n,char a, char b, char c){ if (n == 1) { cout << a << "->" << c << endl; return; } digui(n - 1, a, c, b);

#include<iostream>#include<vector>#include<map>#include<string>#include<cstring>#include<cstdio>#include<algorithm>#include<set>#include<queue>#include<unordered_map>#include<cmath&gt

#include<iostream>#include<vector>#include<map>#include<string>#include<cstring>#include<cstdio>#include<algorithm>#include<set>#include<queue>#include<unordered_map>#include<cmath&g.

#include<iostream>#include<vector>#include<map>#include<string>#include<cstring>#include<cstdio>#include<algorithm>#include<set>#include<queue>#include<unordered_map>#include<cmath&g.

dp 将复杂度降到n平方#include<iostream>#include<vector>#include<queue>#include<stack>#include<string>#include<math.h>#include<algorithm>#include<map>#include<cstring>#include<set>using namespace

递归写法 从上往下#include<iostream>#include<vector>#include<queue>#include<stack>#include<string>#include<math.h>#include<algorithm>#include<map>#include<cstring>#include<set>using namespace s

int 表示的数据范围有限，当涉及大整数的运算时候，我们就只能借助数组来存储，借助数组就依靠手工加减乘除，就小学学的那种来运算，然后一个个存入。先来个结构体bign，含有一个数组和数组长度。我们用字符串读入的数字可以存入bign。头尾颠倒存入，因为模拟手工加减乘这三个都是从低位开始的，颠倒后方便操作bign change(char str[]){ bign a; a.len = strlen(str); for (int i = 0; i < a.len; i++) { a.d[i]

#include<iostream>using namespace std;int Qtoten(int number,int p){ int product=1,x=0; while(number!=0) { x=x+(number%10)*product; number=number/10; product*=p; } return x;}void temtoQ(int number,int p){ int temp[40],count=0,num=0;

#include<iostream>using namespace std;int x[1000] = {0};int main() { x[0] = 1; int i, j, k = 1; int temp = 1; for (i = 2; i <= 50; i++) { for (int j = 0; j < k; j++) { x[j] = i * x[j];//2*1 3*2 4*6 } for (j = 0; j < k; j

#include <iostream>using namespace std;int main(){ int a, b, temp1, temp2; int c; cout << "请按从大到小的顺序两个数字" << endl; cin >> a >> b; temp1 = a, temp2 = b; while (temp2 != 0) { c = temp1 % temp2; temp1 = temp2; te.

void merge(int a[], int l1, int r1, int l2, int r2){ int i = l1, j = l2; int temp[100], index = 0; while (i <= r1 && j <= r2) { if (a[i] <= a[j]) temp[index++] = a[i++]; else temp[index++] = a[j++]; } while (i <= r1) .

为什么要用快速幂因为当指数很大的时候，暴力的求幂函数的方式导致复杂度非常高。因此要采用降低复杂度的方式，称作快速幂。复杂度都是log2N思想二分法思想用递归 比如2的10次方拆成两个2的5次方相乘 2的5次方再拆成2*2的4次方，2的四次方再拆成2的2次方 一直到指数为0。用迭代typedef long long LL;//递归写法LL binarypow1(LL a, LL b, LL m){ if (b == 0) return 1; if (b % 2 == 1) r

[转]https://blog.youkuaiyun.com/flashmsn/article/details/94642687题目描述:给出N个线段长度，试将它们头尾相接组合成一个凸多边形，使凸多边形的外接圆（多边形每个顶点都在圆上）的半径最大，求该最大半径。其中N<=10^5，线段长度均不超过100，要求算法中不涉及坐标的计算。思路：二分算法的本质就是通过不断迭代使left 和 right 在固定条件下逐渐靠近真实值，符合一定误差，所以实际上把该题放在二分扩展里面，这个所谓的最大半径的“最大”是不在求

#include<iostream>#include<cstring>#include<algorithm>#include<string>using namespace std;int main(){ int n, k; cin >> n; int a[1000]; for (int i = 0; i < n; i++) { cin >> a[i]; } cin >> k; sort(.

#include<iostream>#include<cstring>#include<algorithm>#include<string>int n, p[9], hashtable[9] = { 0 }, num = 0;using namespace std;void digui(int index){ if (index == n + 1) { num++; return; } for (int x = 1; x <

#include<iostream>#include<cstring>#include<algorithm>#include<string>int n, p[10], hashtable[10] = { 0 };using namespace std;void digui(int index){ if (index == n + 1) { for (int i = 1; i <=n; i++) { cout <&lt

处理很多输入数据的时候可以边输入边处理，没必要读完了在处理可以简化很多。建立一个很大的桶像桶排序一样往里放数据这样可能更好解决问题。

想了挺久的，scanf读取不完全的问题，所以使用return的时候要注意是否会对下一次循环有影响，return 和break等同效果在这。#include<iostream>using namespace std;int flag=1;int test(int v[4]){ for(int i=0;i<4;i++){ scanf("%d",&v[i]...