咸鱼

#include <iostream>using namespace std;int main(){ int a[100], n; while (cin >> n){ int s = 1; for (int i = 0; i < n; i++) cin >> a[i]; for (int i = 0; i <

#include<iostream>#include<string>using namespace std;int main(){ int n, length; string str; cin >> n; getchar(); while (n){ int num = 0; getline(cin, str);

#include <iostream>using namespace std;int main(){ int n; int a[100]; int x, y, r; int gcd; //greatest common divisor int lcmp; //least common multiple while (cin >> n)

#include <iostream>using namespace std;int main(){ int n, s = 0; int a[100]; while (cin >> n && n){ for (int i = 0; i < n; i++){ cin >> a[i]; s += a[i] / 1

#include <iostream>#include <string>using namespace std;int main(){ int n; cin >> n; getchar(); while (n--){ string str; getline(cin, str); int length = str.l

#include<iostream>#include<string>using namespace std;int main(){ string str; while (getline(cin, str)){ int length = str.length(); //第一个字符是小写就要转换 if (str[0] >= 'a' &

#include <iostream>using namespace std;int main(){ int n,t; int a[100]; while (cin >> n && n != 0){ for (int i = 0; i < n; i++) cin >> a[i]; //感觉就是“冒泡”排序吧，第一轮第一

#include <iostream>using namespace std;int main(){ int n, m,t; int a[100]; //双0结束，这里讲实话有点迷，“或，且”，里面的括号不要也可以AC，有点懒不想去找东西看，先放着。 while (cin >> n >> m && (n != 0 || m != 0)){ for

#include <iostream>#include <string>using namespace std;int main(){ string str; while (getline(cin, str)){ string str2 = str; int length = str.length(); char c = 'a';

#include <iostream>#include <string>using namespace std;int main(){ int n; cin >> n; getchar(); while (n--){ string str; int num = 0; getline(cin, str);

#include <iostream>using namespace std;int main(){ int n; int a[30][30]; while (cin >> n){ for (int i = 0; i < n; i++){ a[i][0] = 1; a[i][i] = 1; }

#include <iostream>using namespace std;int main(){ int n; cin >> n; while (n){ int M; cin >> M; long int a[40]; a[0] = 0; a[1] = 1; for (in

其实是个很简单的题目，我自己把自己绕进去了，以为要实现大数相乘，然后保留下后三位数- -! 题目要求保留最后三位数，那就只求最后三位数跟大数的乘积 看网上高端的说法是 同余定理 数论里面的概念 有点绕- -! 额，顺带说下，这道题好像是道挺水的题，有写这道题的博客都好少- -!#include <iostream>using namespace std;int main(){ i

开始没明白真约数是啥- -！贼简单。#include <iostream>using namespace std;int find_sum(int& n){ int k = n; int sum = 1; for (int i = 2; i <k; i++){ if (k % i == 0){ sum += i; }

很坑很坑，数据类型不能是int，不能是int，int….. 我还以为两边和大于第三边这个思路(两边差小于第三边，这两个只要考虑一个就可以了)是错的，错的，错的！ 还记得某次上机考试也是考的这个，血亏血亏血亏！！！！！#include <iostream>using namespace std;struct trig{ double a; double b; doubl

思路：将电视节目按结束时间从小到大排列，越早结束，越能看更多的节目； 以结束时间顺序找上个节目结束后能开始的节目，统计个数； 额，好像是贪心算法的一个简单题，没到那个层次，作为小白的我任重而道远啊- -!。#include <iostream>using namespace std;struct program{ int start_time; int end_time;}

//A集合相对于B集合的差集#include <iostream>using namespace std;int main(){ int A[100], B[100], C[100]; int n, m; //第一次提交，(n != 0 && m != 0)，&& 和 || 一段时间不用不记得- - while (cin >> n >> m && (n != 0

//读取文件数据#include <iostream>#include <string>#include <fstream>using namespace std;int main(){ string b[100]; string file_path = "D:\\P1.txt"; //cout << file << endl; ifstream infile

#include <iostream>using namespace std;int main(){ int n; int ah, am, as, bh, bm, bs; int s, m, h; cin >> n; while (n--){ cin >> ah >> am >> as >> bh >> bm >> bs;

#include <iostream>#include <string>using namespace std;int main(){ int n, flag = 0,length; string a; cin >> n; getchar(); while (n){ getline(cin, a); length = a.

#include <iostream>using namespace std;int main(){ int n, m; while (cin >> m >> n){ signed int a[100][100]; signed int max = 0; int x, y; //输入m*n数据 for

#include <iostream>using namespace std;int main(){ double sum = 0, avr; int n,max=0,min=100; int a[100]; while (cin >> n){ for (int i = 0; i < n; i++){ cin >> a[i]

#include <iostream>using namespace std;int Fun(int a){ int f, n = a; if (n == 2) f = 4; else f = (Fun(n-1) + 1) * 2; return f;}int main(){ int n,S; while (cin >> n){

#include <iostream>#include <cmath>using namespace std;int isPrime(const int a){ int flag = 1; int b = a; for (int i = 2; i < sqrt(b); i++){ if (b%i == 0) flag=0; } retur

#include <iostream>using namespace std;int main(){ int m; int a[1000]; cin >> m; for (int i = 0; i < m; i++) cin >> a[i]; double s = 0, t; for (int i = 0; i < m; i++){

#include <iostream>using namespace std;int main(){ int m, n; int a, b, c,flag=0; int x[1000]; while (cin >> m >> n) { if (m > n){ n = m + n; m = n

#include <iostream>#include <cmath>using namespace std;int main(){ double n, m; while (cin >> n >> m){ double s = n; for (int i = 1; i < m; i++){ //从第二项起递增开方；m-1次

#include <iostream>using namespace std;int main(){ double x[100]; int n,a=0,b=0,c=0; while (cin >> n && n != 0){ for (int i = 0; i < n; i++) cin >> x[i]; for (

#include <iostream>using namespace std;int main(){ long int x, y; int a[1000]; while (cin >> x >> y){ if (x > y){ //大小数交换位置； x = x + y; y = x - y;

//输出格式出错#include <iostream>using namespace std;int main(){ int n, m, y, x, sum = 0, ave; int a[100]; while (cin >> n >> m){ x = n / m; y = n % m; for (int i = 0;

#include <iostream>using namespace std;int main(){ int n, min,t=0,temp; int a[100]; while (cin >> n && n){ for (int i = 0; i < n; i++) cin >> a[i]; min = a[0]; for

#include <iostream>using namespace std;int main(){ int N, R; while (cin >> N >> R){ int a[10], i = 0, flag = 0; int x = N, y; //考虑负值 if (x < 0){ x

#include <iostream>using namespace std;//开始错误的递归想法，f(n)=n,n<=4;f(n)=f(n-1)+n-3,n>4;//正确的递归应该是这样的：f(n)=f(n-1)+第n年新母牛数，而第n年只有第n-3年的母牛能生母牛，即f(n)=f(n-1)+f(n-3)。int f(const int x){ int y = x,s;

杭电ACM2001#include <iostream>#include <cmath>using namespace std;int main(){ double x1,y1,x2,y2; while (cin >> x1 >> y1 >> x2 >> y2){ cout.precision(2); cout << fixed << sqrt(

#include <iostream>#define PI 3.1415927using namespace std;int main(){ double r,V; while (cin >> r){ V = 4/3*PI*r*r*r; cout.precision(3); cout << fixed << V << endl;

#include <iostream>using namespace std;int main(){ int year, month, day,s; char c; while (cin >> year >> c >> month >> c >> day){ switch (month){ case 1:s = 0; break;

#include <iostream>using namespace std;int main(){ int s; while (cin >> s){ if (s >= 90 && s <= 100)cout << "A" << endl; else if (s >= 80 && s <= 89)cout << "B" << endl;

#include <iostream>using namespace std;int main(){ int n, count = 0; char a[100]; cin >> n; while (n){ cin >> a; for (int i = 0; a[i] != '\0'; i++){ if (a[

#include <iostream>using namespace std;int main(){ double a; while (cin >> a){ if (a < 0)a = -a; cout.precision(2); cout << fixed << a << endl; } return 0;}

不能用劳伦公式求面积，没有考虑到凹多边形的情况。 以原点为三角形一个顶点，找多边形上逆时针相方向邻的两个点为依次为三角形的另两个顶点，用叉乘法求三角形面积。 叉乘求面积是有方向，即正负的，但逆时针选点所求面积为正。 逆时针选点的终点为首尾相连。#include <iostream>using namespace std;struct points{ int x; int y