Digit sum --- hit the table a small expert

Portals
are given a number of groups, the number n and a number a b
b is a decimal number to turn, find the 1 - n b are converted to the binary number and each of

Timeout version

#include <iostream>
#include <cstdio>
#define rep(i,a,b) for(int i=a;i<b;i++)
using namespace std;
int n,b;
int change(int x){
    int ans=0;
    while(x){
        ans+=x%b;
        x/=b;
    }
    return ans;
}
int main(){
    int t;
    scanf("%d",&t);
    rep(i,1,t+1){
        scanf("%d%d",&n,&b);
        int ans=0;
        rep(i,1,n+1){
            ans+=change(i);
        }
        printf("Case #%d: %d\n",i,ans);
    }
    return 0;
}

Play table thinking

For t enter the words, but the only answer to each value can be used to play table

#include <iostream>
#include <cstdio>
#define rep(i,a,b) for(int i=a;i<b;i++)
using namespace std;
const int N=1e6+5;
int a[N][15];
int main(){
    rep(i,2,11){
        rep(j,1,N){
            int t=j,cnt=0;
            while(t){
                cnt+=t%i;
                t/=i;
            }
            a[j][i]=a[j-1][i]+cnt;
        }
    }
    int t;
    scanf("%d",&t);
    rep(i,1,t+1){
        int n,b;
        scanf("%d%d",&n,&b);
        printf("Case #%d: %d\n", i,a[n][b]);
    }
    return 0;
}