liuxiaocs7的博客

最开始基础的A + B problem关键在于Process to end of file 就是处理到文件的结束此处为坑#include <iostream>using namespace std;int main(){ int a; int b; while(cin>>a>>b) { cout<<(a+b)<&lt...

#include <iostream>#include <cstdio>#include <cmath>using namespace std;bool IsPrime(int x){ int len = (int)sqrt(x); for(int i=2; i<=len; i++) { if(x%i == 0) retu...

迭代#include <iostream>using namespace std;int main(){ int n; while(cin >> n) { int today = 1; int yesterday = 0; for(int i=1; i<n; i++) { yesterday = (today+1)*2; ...

输入结束条件为0；

#include <iostream>#include <cstdio>using namespace std;int main(){ int n; while(cin >> n) { int sum = 0; int a; cin >> a; int max = a; int min = a; sum +...

#include <iostream>using namespace std;int main(){ int n, m; while(cin >> n >> m) { int sum = 0; int a[n+1] = {0}; bool flag = true; for(int i=1; i<=n; i++) {...

输入结束条件为n = m = 0输入格式很重要先输入的整数代表输入块结果是计算每一个块中的每一组测试用例的结果。整体思路是二重循环遍历所有情况 –判断一个数是否为整数bool isInt(double x){ long int y=(long int) x; //if ((x-y)!=0)//这段代码可能会丢失精度 if((x-y)<=0.00000...

#include <iostream>#include <algorithm>using namespace std;int main(){ int n; while(cin >> n && n!=0) { int a[n]; for(int i=0; i<n; i++) { cin >> ...

#include <iostream>#include <algorithm>using namespace std;int main(){ int n; cin >> n; while(n--) { string s; cin >> s; int len = s.size(); int sum = 0; f...

类似于斐波那契数列。递推式:f(n) = n (n<=4)f(n) = f(n-1) + f(n-3) (n>4)递归解决：#include <iostream>using namespace std;int sum(int n){ if(n<=4) { return n; } else { return sum(n-1)+...

#include <iostream>#include <cstdio>#include <cmath>using namespace std;int main(){ int n; while(cin >> n && n!=0) { //int a[n]; int *a = new int[n]; f...

多个测试用例#include <iostream>using namespace std;int main(){ int n; // 老师数 int sum = 0; // 总计钞票的数量 int m; int x; // 每个老师的钱 int a[6] = {100, 50, 10, 5, 2, 1}; //存放币值 while(true) {...

一开始将输入的数据存入数组中，一直爆掉#include <iostream>#include <cmath>using namespace std;int main(){ int m, n; while(true) { cin >> m>> n; long **s = new long* [m]; for(int i=0...

#include <iostream>#include <cstdio>using namespace std;int main(){ int m; cin >> m; while(m--) { int n; cin >> n; double sum = 0.0; for(int i=1; i<=n; i...

#include <iostream>using namespace std;bool IsFlower(int n){ int sum = 0; int m = n; while(n) { sum += (n%10)*(n%10)*(n%10); n /= 10; } if(sum == m) { return true; } else ...

#include <iostream>#include <cmath>#include <cstdio>using namespace std;int main(){ double n; int m; while(cin >> n >> m) { double sum = 0; while(m--) ...

#include <iostream>using namespace std;int main(){ int a; while(cin>>a) { int sum = 0; for(int i=1; i<=a; i++) sum += i; cout<<sum<<endl<<endl; }...

  #include <iostream>using namespace std;int a, b;int process(int n);int main(){ int n; while(1) { cin>>a>>b>>n; if(a==0 && b ==0 && n==0)...

思路:外层循环控制问题数，内层读入输入计算最小公倍数求最小公倍数=两数的乘积/最大公约数错误。。。#include <iostream>using namespace std;int LeastNum(int a, int b){ int c = 0; int m = a; //保存a,b的值 int n = b; while(b != 0) { c =...

斐波那契数列的变形而已。#include <iostream>using namespace std;int Fbi(int n){ if(n == 0) { return 11; } else if(n == 1) { return 7; } else { return Fbi(n-1) + Fbi(n-2); }}int main...

思路读入总共的层数(确保层数不为0，否则直接退出循环)层数为接下来循环的次数，每次读入楼层，注意每一次读入新的楼层数，意味着在该楼层都要停留5秒钟，即time += 5;为必要操作接下来判断读入楼层与当前楼层之间的关系，分别对应上楼和下楼两种情况，循环即将结束时，更新当前楼层。输出时间即处理完一次。。。#include <iostream>using namesp...

AC代码思路主要是注意输出的格式问题，以及求e时连续的九项输出，可以边累加边输出，不必要每次计算载输出。#include <iostream>#include <cstdio>using namespace std;int factorial(int n);int main(){ cout<< "n " << "e"...

#include <iostream>using namespace std;char s[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p','q','r', 's','t','u','v','w','x','y','z'};int main()...

直接AC#include <iostream>#include <cstdio>#include <cmath>using namespace std;int main(){ double x1, y1, x2, y2; while(cin >> x1 >> y1 >> x2 >> y2) {

#include <iostream>#include <cstdio>#define PI 3.1415927using namespace std;int main(){ double r; while(cin >> r) { double V = 4*PI*r*r*r/3; printf("%.3f\n", V); } ...

#include <iostream>#include <cstdio>#include <cmath>using namespace std;int main(){ double r; while(cin >> r) { printf("%.2f\n", abs(r)); } return 0;}

#include <iostream>#include <cstdio>#include <cmath>using namespace std;int main(){ int score; while(cin >> score) { if(score<0 || score>100) { cout &lt...

#include <iostream>#include <cstdio>#include <cmath>bool IsLeapYear(int year){ return ((year%4==0&&year%100!=0) || year%400==0);}using namespace std;int main(){...

#include <iostream>using namespace std;int main(){ int n; while(cin >> n) { int result = 1; while(n--) { int x; cin >> x; if(x % 2 == 1) { result *= x;...

输入数据没有判断x和y的大小。。。#include <iostream>using namespace std;int main(){ int x, y; while(cin >>x >> y) { int sum1 = 0; int sum2 = 0; for(int i=x; i<=y; i++) { if(i...

#include <iostream>using namespace std;int main(){ int n; while(cin >> n && n!=0) { int n1 = 0; int n2 = 0; int n3 = 0; while(n--) { double x; cin >>...