近似模拟_百炼 计算对数 2739(换底公式的应用)

 1 #define _CRT_SECURE_NO_WARNINGS  
 2 #include <stdio.h>
 3 #include <math.h>
 4 #include <algorithm>
 5 #include <stdlib.h>
 6 #include <vector>
 7 #include <map>
 8 #include <queue>
 9 #include <string>
10 #include <iostream>
11 #include <ctype.h>
12 #include <string.h>
13 #include <set>
14 #include <stack>
15 #include<functional>
16 using namespace std;
17 #define Size 105
18 #define maxn  1<<30
19 #define minn  1e-6
20 char str1[Size], str2[Size];
21 
22 int main(){
23     memset(str1, 0, sizeof(str1));
24     memset(str2, 0, sizeof(str2));
25     scanf("%s%s", str1, str2);
26     int len1 = strlen(str1);
27     int len2 = strlen(str2);
28     for (int i = 0; i < 6; i++)
29     {
30         if (str1[i] == ‘\0‘) str1[i] = ‘0‘;
31         if (str2[i] == ‘\0‘) str2[i] = ‘0‘;
32     }
33     str1[6] = str2[6] = 0;
34     /*
35     求loga(b) = log10b/log10a
36     log10b 的近似值是 lenstrb-1+log10(atoi(str2) / 1e5)  后面的那么多位数不用看了
37     */
38     printf("%d\n", int((log10(atoi(str2) / 1e5) + len2 - 1) / (log10(atoi(str1) / 1e5) + len1 - 1)));
39     //注意是除过以后再取整而不是取整以后再除
40     system("pause");
41 }