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来自洛谷，原作者为

	wangyanjing **

搬运于2025-08-24 22:05:48，当前版本为作者最后更新于2025-01-14 14:14:05，作者可能在搬运后再次修改，您可在原文处查看最新版

自动搬运只会搬运当前题目点赞数最高的题解，您可前往洛谷题解查看更多

以下是正文

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前置知识

$$\log_a(b) = {\log(b) \over \log(a)}$$ $$\log(ab) = \log(a) + \log(b)$$ $$\log_c(ab) = \log_c(a) + \log_c(b)$$ $$len = \log_{10}(n) + 1$$

题意

为了方便书写，这里的 $s_i$ 为题目给出的数的第 $i$ 位。

题意可简化为求以下式子的值：

$$\log_{10}(\prod_{i = 0}^{n-1} a^i \cdot \prod_{i = 1}^n s_i)+1 $$

Solution

$$\log_{10}(\prod_{i = 0}^{n-1} a^i \cdot \prod_{i = 1}^n s_i)+1 $$$$= \log_{10}(a^{\sum_{i = 0}^{n-1} i} \cdot \prod_{i = 1}^n s_i)+1 $$$$= \log_{10}(a^{n(n-1) \over2} \cdot \prod_{i = 1}^n s_i)+1 $$$$= \log_{10}(a^{n(n-1) \over2}) + \log_{10}(\prod_{i = 0}^n s_i)+1 $$$$= \log_{10}(a^{n(n-1) \over2}) + \sum_{i = 1} ^n \log_{10} s_i +1 $$$$= {\log(a^{n(n-1) \over2}) \over \log (10)} + \sum_{i = 0} ^n \log_{10} s_i +1 $$$$= {\log(a^{n(n-1) \over2}) \over \log (10) \cdot \log(a)} \cdot \log(a) + \sum_{i = 1} ^n \log_{10} s_i +1 $$$$= {(n-1)n \cdot \log(a) \over 2\log (10)} + \sum_{i = 1} ^n \log_{10} s_i +1 $$$$= {(n-1)n \over 2} \cdot \log_{10} a + \sum_{i = 1} ^n \log_{10} s_i +1 $$

Code

#include<cstdio>
#include<cmath>
#include<cstring>
const int N = 1e5+5;
char S[N];
#define ll long long
inline void Solve(void){
	int a;
	std::scanf("%d",&a);
	std::scanf("%s",S+1);
	int n = std::strlen(S+1);
	long double ans = 0;
	for(int i = 1;i<=n;++i) ans += std::log10(S[i]-'0');
	ans += 1.0 * n * std::log10(a) / 2 * (n-1);
	ll len = std::floor(ans) + 1ll;
	std::printf("%lld\n",len);
}
int main(){
	int T;
	std::scanf("%d",&T);
	while(T--) Solve();
	return 0;
}