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All in one 1413: determine base
2022-07-05 04:10:00 【Studying hard】
【 Title Description 】
6×9=42 It's wrong for the decimal system , But for 13 It's correct in hexadecimal . namely , 6(13)× 9(13)= 42(13), and 42(13)=4×131+2×130=54(10).
Your task is to write a program , Read in three integers p、q and r, Then determine a base B(2<=B<=40) bring p × q = r. If B There are many options , Output the smallest one .
for example :p=11, q=11, r=121. Then there are 11(3)× 11(3)= 121(3) because 11(3)= 1 × 31+ 1 × 30= 4(10) and 121(3)=1×32+2×31+1×30=16(10). For hexadecimal 10, There are also 11(10)× 11(10)= 121(10). In this case , The output should be 3. If there is no suitable base , The output 0.
【 Input 】
a line , Contains three integers p、q、r. p、q、r All bits of are numbers , also 1 ≤ p、q、r ≤ 1,000,000.
【 Output 】
An integer : Even if have to p×q=r The smallest established B. If there is nothing suitable B, The output 0.
【 sample input 】
6 9 42
【 sample output 】
13
#include<stdio.h> #include<math.h> int getmin(int k,int j) { int temp; int i = 0; int count = 0; while (1) { temp = k % 10; count += temp * pow(j, i); i++; k /= 10; if (k == 0) { break; } } return count; } int main() { int m, n, k; scanf("%d%d%d", &m, &n, &k); int flag = 1; int num1, num2, num3; int j; for (j = 2; j <= 40; j++) { num1 = getmin(m, j); num2 = getmin(n, j); num3 = getmin(k, j); if (num1 * num2 == num3) { flag = 0; break; } } if (flag == 1) { printf("0"); } else { printf("%d", j); } return 0; }
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