1+2+3+..+100
难度 0
#include<iostream>
using namespace std;
int main()
{
int i, s = 0;
// 第一种
for (i = 0; i <= 100; i++)
{
s += i;
}
cout << s << endl;
// 第二种
s = i = 0;
while (i < 100)
{
i++;
s += i;
}
cout << s << endl;
// 第三种
s = i = 0;
do
{
i++;
s += i;
}
while (i < 100);
cout << s << endl;
// 第四种(不推荐)
s = i = 0;
st:
i++;
s += i;
if (i < 100)
goto st;
cout << s << endl;
return 0;
}

阶乘连加
难度1
#include<iostream>
using namespace std;
int main(){
int s=0, i=1, t, n;
cout << "输入n ";
cin >> n;
for (t=1;t<=n;t++){
i *= t;
s += i;
}
cout << "1!+2!+3!+...+" << n
<< "! = " << s << endl;
return 0;
}

闰年平年
难度1
#include<iostream>
using namespace std;
int main(){
int year;
cout<<"input a year:";
cin>>year;
if((year%4==0 && year%100!=0) || year%400==0) cout<<"yes\n";
else cout<<"no\n";
return 0;
}

斐波那契数列
难度2
#include <iostream>
using namespace std;
int f(int x){
if (x==1 or x==2) return 1;
else return f(x-1)+f(x-2);
}
int main(){
cout << f(30);
return 0;
}

冒泡排序
难度2
#include<iostream>
using namespace std;
int main()
{
int o, i;
int t;// 容器
int work[10] = { 0, 55, 65, 68, 25, 38, 67, 51, 62, 98 };
for (o = 0; o <= 8; o++)
{
for (i = o + 1; i <= 9; i++)
{
if (work[o] < work[i])
{
t = work[o];
work[o] = work[i];
work[i] = t;
}
}
}
for (i = 0; i <= 9; i++)
{
cout << work[i] << " ";
}
return 0;
}

冰雹猜想
难度2
#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
while (1){
unsigned long int a;
cin >> a;
while (1){
if ((a % 2) == 0)
a /= 2;
else
a = a*3+1;
cout << a << endl;
if (a==1)
break;
}
cout << "end" << endl;
}
return 0;
}

支持运算符优先级的四则运算
难度3
#include<iostream>
#include<vector>
using namespace std;
int main(){
vector<double> add;
double i;
char f='+';
bool first=true;
while (1){
if (!first)
cin >> f;
if (f!='=')
cin >> i;
else
break;
switch (f){
case '+':
add.push_back(i);
break;
case '-':
add.push_back(-i);
break;
case '*':
add[add.size()-1]*=i;
break;
case '/':
add[add.size()-1]/=i;
break;
}
first=false;
}
i = 0;
for (int t=0; t<add.size(); t++)
i += add[t];
cout << i;
return 0;
}

奇数阶幻方
难度3
#include<iostream>
#include<iomanip>
using namespace std;
int main(){
int n;
cin >> n;
int a[n+1][n+1];
int i, j;
int t;
int k, l;
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
a[i][j]=0;
i=1;
j=n/2+1;
a[i][j]=1;
for (t=2;t<=n*n;t++){
k=i-1;
l=j+1;
if (k==0 && l>n){
k+=2;
l=j;
}
else if (k==0 && !(l>n))
k=n;
else if (k!=0&&l>n)
l=1;
else if (a[k][l]!=0){
k+=2;
l=j;
}
a[k][l]=t;
i=k;
j=l;
}
for (i=1;i<=n;i++){
for (j=1;j<=n;j++){
cout <<setw(4)<< a[i][j] << " ";
}
cout << endl;
}
return 0;
}

汉诺塔
难度4
#include <iostream>
using namespace std;
void move(int n,char a,char c,char b){
//a到c，用b柱过渡,n个盘
if (n==1){
cout << a << "->" << c << endl;
//如果只一个，直接到目的地
}
else{
//否则先将n-1一个移动到过渡上
/*
在这里，递归执行下去，直到到
达上面的递归边界，再返回
最终完成所有移动
*/
move(n-1,a,b,c);
//输出这一步的路径
cout << a << "->" << c << endl;
//从过渡上过去目的地
move(n-1,b,c,a);
}
}
int main(){
int n;
cin >> n;
//输入盘总数
move(n,'a','c','b');
//把n个盘子从a移到c借助b过渡
return 0;
}

折半查找
难度4
#include <iostream>
#include <cmath>
using namespace std;
int a[100];
int i;
void init(){
cin >> i;
for (int j=1; j<=i; j++)
cin >> a[j];
}
int find (int x, int l, int r){
//cout << l << " " << r << endl;
if (l==r){
if (x == a[l])
return l;
else
return -1;
//找不到
}
int mid=(l+r)/2;
if (a[mid] == x)
return mid;
else if (a[mid] < x)
return find(x, mid+1, r);
else
return find(x, l, mid-1);
}
int main(){
init();
int k;
cin >> k;
cout << find(k,1,i);
return 0;
}

为什么喂我？

第一，我看不懂，特别是cout,com,<<什么的
第二，当我看到冒泡二星的时候就已放弃继续看