#include <bits/stdc++.h>
using namespace std;
int main()
{
freopen("2015coin.in","r",stdin);
freopen("2015coin.out","w",stdout);
int n,jinbi=0,days=0;
cin>>n;
for(int i = 1;i<=n;i++)
{if(days>=n) break;
for(int j = 1;j<i+1;j++)
{jinbi+=i;
days++;
if(days>n) jinbi-= i;}}
cout<<jinbi;}

简单

#include<bits/stdc++.h>
using namespace std;
int main(){
long long a,j=0;
freopen("prime.in","r",stdin);
freopen("prime.out","w",stdout);
cin>>a;
for(long long int i=2;i<sqrt(a);i++)
{if(a%i==0) {cout<<a/i; break;fclose(stdin);fclose(stdout);}}return 0;}

诶亚，这怎么回事儿啊这是

不嘻嘻

12345678900987654321234

#include<bits/stdc++.h>
using namespace std;
int main()
{freopen("twoj.in","r",stdin );
freopen("twoj.out","w",stdout );
int l,r,j,x=0,k=0;
cin>>l>>r;
for(int i=l;i<=r;++i)
{j=i;while(j)
{
k=j%10;
if(k==2) x++;
j=j/10 ;}}
cout<<x;
return 0;
fclose(stdin);fclose(stdout);}

多谢前辈

#include<bits/stdc++.h>
using namespace std;
int main()
{freopen("count2013.in","r",stdin );
freopen("count2013.out","w",stdout );
int n,z,j,x=0,k=0;
cin>>n>>z;
for(int i=1;i<=n;++i)
{j=i;while(j)
{
k=j%10;
if(k==z) x++;
j=j/10 ;}}
cout<<x;
return 0;
fclose(stdin);fclose(stdout);}

回复 @2018noip必胜！ :
#include<bits/stdc++.h>
using namespace std;
int main(){
int a,b=0;
freopen("reverse.in","r",stdin);
freopen("reverse.out","w",stdout);
cin>>a;
while (a)
{b=b*10+a%10;
a/=10;}
cout <<b;}
。。。。。。。。。

回复 @father :
just so so

so easy

萌新表示不会

都有实力

#include<bits/stdc++.h>
using namespace std;
long long b,p,k;
int f(int a)
{
if(a==0) return 1;
int t=f(a/2)%k;
t=(t*t)%k;
if(a%2==1) t=(t*(b%k))%k;
return t;
}
int main()
{
freopen("dmod.in","r",stdin);
freopen("dmod.out","w",stdout);
cin>>b>>p>>k;
int b1=b;
b%=k;
printf("%d\n",f(p));
}
我又怎么错了?!!!

terrific

回复 @ムラサメ : 大佬你人好好Q^Q

指针数组我们开的大小为 $d_x$ 时，则该数组下标也是从 0 开始的，不能访问 $f[x,d_x]$

喜报: unr d1t3 出现这个模型，被取之。回旋镖哦哦哦哦

幽默topcoder