#include
#define N 100000001
using namespace std;
long long *sum=new long long [N];
int main()
{
int a,n;
cin>>a>>n;
if(n==0) cout<<'0'<<endl;
else {
int T=n;
for(int i=1;i<=n;i++)
{
sum[i]+=a*T;
--T;
int cnt =sum[i]/10;
if(sum[i]>9) sum[i]%=10;
sum[i+1]+=cnt;
}
int t=n*(n+1)/2;
while(1)
{
if(sum[t]!=0) break;
--t;
}
for(int i=t;i>=1;i--)
cout<<sum[i];
cout<<endl;}
delete []sum;
return 0;
}
给定某数字A(1≤A≤9)以及非负整数N(0≤N≤100000),求数列之和S=A+AA+AAA+⋯+AA⋯A(N个A)。例如A=1, N=3时,S=1+11+111=123