//2010
#include<stdio.h>
#include<math.h>
int flower(int a)
{
int sum;
int temp;
temp = a;
sum = 0;
while (a)
{
sum = sum + (a % 10)*(a % 10)*(a % 10);
a /= 10;
}
if (temp==sum)
{
return 1;
}
else
{
return 0;
}
}
int main()
{
int m, n;
int count1 = 0;//记录非水仙花数
int count2 = 0;//记录水仙花数
while (~scanf("%d%d",&m,&n))
{
count1 = 0;
count2 = 0;
for (int i = m; i <= n; i++)
{
if (!flower(i))
{
count1++;
}
else
{
count2++;
if (count2==1)
{
printf("%d", i);
}
else
{
printf(" %d", i);
}
}
}
if (count1==(n-m+1))
{
printf("no");
}
printf("
");
}
return 0;
}
//2011
#include<stdio.h>
#include<math.h>
int main()
{
int m;
int i;
int n;
int j;
double sum ;
while (~scanf("%d",&m))
{
for (i = 1; i <= m; i++)
{
sum = 0;
scanf("%d", &n);
for ( j = 1; j <= n; j++)
{
if (j%2==1)
{
sum = sum + (1.0 / j);
}
else
{
sum = sum - (1.0 / j);
}
}
printf("%.2lf
", sum);
}
}
return 0;
}
//2012
#include<stdio.h>
#include<math.h>
int isPrime(int a)
{
int i;
//是从i=2开始，原来写成是1开始了，就wa了
for (i = 2; i <= sqrt(double(a)); i++)
{
if (a%i==0)
{
break;
}
}
if (i>sqrt(double(a)))
{
return 1;
}
return 0;
}
int main()
{
int x, y;
int result;
int i;
int count;
while (~scanf("%d%d",&x,&y))
{
if (x==0&&y==0)
{
break;
}
result = 0;
count = 0;
//for ( i = x+1; i < y; i++)
//按照题目的输入和输出，输出的是OK，如果是（0，1）范围内的话是没有整数的，何来对应的素数，所以题目应该理解为[0,1]->[x,y]
for (i = x; i <= y; i++)
{
result = i*i + i + 41;
if (isPrime(result))
{
count++;//记录使表达式的值为素数的数的个数
}
}
//if (count==(y-x-1))
if (count==(y-x+1))
{
printf("OK
");
}
else
{
printf("Sorry
");
}
}
return 0;
}
//2013
//思路：设第n天剩下f（n），第n-1天剩下f（n-1），则f（n）= [f(n-1)-(f(n-1)/2+1)],整理得f(n-1)=2*[f(n)+1];但是已知的是第n天剩下的个数，所以应该从后往前推。
//即：第n天是1；第n-1天是2（1+1）；第n-2天是2（1+4）；第n-3天是2（1+10）；...则第一天是2（1+x）[x代表第二天的]
#include<stdio.h>
#include<math.h>
int main()
{
int n;
int i;
int sum;
while (~scanf("%d",&n))
{
sum = 1;
for (i = 1; i < n; i++)
{
sum = 2*(1 + sum);
}
printf("%d
", sum);
}
return 0;
}
//2014
#include<iostream>
#include<algorithm>
#include<iomanip>
#define N 101
double score[N];
using namespace std;
int main()
{
int i;
int n;
double average;
double sum;
while (cin>>n)
{
average = 0;
sum = 0;
for ( i = 0; i < n; i++)
{
cin >> score[i];
}
sort(score, score + n);
for ( i = 1; i < n-1; i++)
{
sum += score[i];
}
average = sum / (n - 2);
//保留两位小数
cout << fixed<< setprecision(2)<< average << endl;
}
return 0;
}
//2015
#include<iostream>
#include<algorithm>
#include<iomanip>
#define N 101
int num[N];
using namespace std;
int main()
{
int n, m;
int j;
int i;
num[0] = 0;
int average;
int sum;
int cnt1;
int cnt2;//用来记录平均数的个数
for (i = 1; i <= 100; i++)
{
num[i] = i * 2;
}
while (cin>>n>>m)
{
average = 0;
sum = 0;
cnt1 = 0;
for (i = 1; i <= n; i++)
{
//average = 0;
sum += num[i];
cnt1++;//用来记录是否是达到m
/*if (cnt2)
{
cout<<
}*/
if (cnt1==m&&i!=n)
{
cout << sum / cnt1<<' ';
cnt1 = 0;
sum = 0;
}
/*for ( j = i; j < i + m,j<=n; j++)
{
sum = sum + num[j];
}
average = 1.0*sum / m;
if (i==1)
{
cout << average;
}
else
{
cout << ' ' << average;
}*/
}
if (i==n+1)
{
cout <<sum/(cnt1)<< endl;
}
}
return 0;
}
//2016
#include<iostream>
#include<algorithm>
#include<iomanip>
#define N 101
int num[N];
using namespace std;
int main()
{
int n;
int i;
int min;
int min_pos;
int temp;
while (cin>>n)
{
if (n==0)
{
break;
}
cin >> num[0];
min = num[0];
min_pos = 0;
for ( i = 1; i < n; i++)
{
cin >> num[i];
if (min>num[i])
{
min = num[i];
min_pos = i;
}
}
//交换最小数和第一个数
temp = num[min_pos];
num[min_pos] = num[0];
num[0] = temp;
for ( i = 0; i < n; i++)
{
if (i == 0)
cout << num[i];
else
{
cout << ' ' << num[i];
}
}
cout << endl;
}
return 0;
}
//2017
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<iomanip>
#include<string.h>
#define N 101
char num[N];
using namespace std;
int main()
{
int n;
int count;
//int len;
int i;
while (~scanf("%d",&n))
{
while (n--)
{
count = 0;//记录数字的个数
//gets(num);
scanf("%s", num);
for (i = 0; i < strlen(num);i++)
if (num[i]>='0'&&num[i]<='9')
{
count++;
}
//cout << count << endl;
printf("%d
", count);
}
}
return 0;
}
//2019
//思路：关键是找到插入的位置
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<iomanip>
#include<string.h>
#define N 101
int num[N];
using namespace std;
int main()
{
int n;
int m;
int i;
int pos;//插入的位置
while (cin>>n>>m)
{
if (n==0&&m==0)
{
break;
}
pos = 0;
for ( i = 0; i < n; i++)
{
cin >> num[i];
}
for ( i = 1; i < n; i++)
{
//if (m>num[i - 1] && m < num[i])
if (m<num[i])
{
pos = i;//插入在下标为i-1后面
break;
}
}
//网上的想法：直接打印前部分的，接着打印m,最后打印后部分的
for ( i = 0; i < pos; i++)
{
cout << num[i] << ' ';
}
cout << m;
for ( i = pos; i < n; i++)
{
cout << ' ' << num[i];
}
//我的想法：因为m要插入到数组中，所以后面的要向后移动一个
/*for ( i = n-1; i >= pos+1; i--)
{
num[i+1] = num[i];
}
num[pos + 1] = m;
cout << num[0];
for ( i = 1; i < n+1; i++)
{
cout << ' ' << num[i];
}*/
cout << endl;
}
return 0;
}