Codeforces Round #747 (Div. 2)B. Special Numbers

Posted 2022-07-12 ~晚风微凉~

1.题目：

Theofanis really likes sequences of positive integers, thus his teacher (Yeltsa Kcir) gave him a problem about a sequence that consists of only special numbers.

Let’s call a positive number special if it can be written as a sum of different non-negative powers of n. For example, for n=4 number 17 is special, because it can be written as 40+42=1+16=17, but 9 is not.

Theofanis asks you to help him find the k-th special number if they are sorted in increasing order. Since this number may be too large, output it modulo 109+7.

Input
The first line contains a single integer t (1≤t≤104) — the number of test cases.

The first and only line of each test case contains two integers n and k (2≤n≤109; 1≤k≤109).

Output
For each test case, print one integer — the k-th special number in increasing order modulo 109+7.

Example
inputCopy
3
3 4
2 12
105 564
outputCopy
9
12
3595374
Note
For n=3 the sequence is [1,3,4,9…]

*思路：*感觉这个可以推出一个公式出来，不管是是哪个数字
想到了从十进制推到二进制数的形式。

2.官方题解思路如下：

nice！和想的一样，但是做不到把思路复现出来

3.代码部分：

#include<bits/stdc++.h>
typedef long long ll;
const ll mod=1e9+7;//注意不能写成#define mod 1e9+7,这样编译会在第21,23报错，lld%int
using namespace std;
int t;

int main()

    scanf("%d",&t);
    while(t--)
    
        int n,k;
        ll ans=0;
        ll p=1;
        scanf("%d %d",&n,&k);
        for(int i=0;i<31;i++)
        
            if(k &(1<<i))
            
               ans = (ans + p ) % mod;//就像转化为二进制位，加入此为为1则加上对应的进制值
            
            p = p*n;//为0，继续累乘，但不加入答案中
            p = p % mod;
        
        printf("%lld\\n",ans);
    
    return 0;