$n = a_0 + a_1 + a_2 + … + a_k$
$m = a_0(n+1)^0 + a_1(n+1)^1 + a_2(n+1)^2 + \dots + a_k(n+1)^k$

m视为(n+1)进制 进制转换取得$a_0 \dots a_k$

#include<bits/stdc++.h>

using namespace std;
const int N = 1e6+10;
int a[N];

int main()
{
    int n,m; scanf("%d %d",&n,&m);
    int res = m; int l = 0;
    while(res)
    {
        // printf("%d\n",res%11);
        a[l++] = res % (n+1);
        res = res / (n+1);
    }
    for(int i=l-1;i>=0;i--)
    {
    //  printf("%d\n",a[i]);
        if(a[i]>1&&i>1){
            printf("%dx^%d+",a[i],i);
        }else if(a[i] == 1&&i>1){
            printf("x^%d+",i);
        }else if(a[i] > 1 && i == 1){
            printf("%dx+",a[i]);
        }else if(a[i] == 1 && i == 1){
            printf("x+");
        }

        if(a[i]>=1&&i == 0){
            printf("%d\n",a[i]);
        }
    }

    return 0;
}