113 - Power of Cryptography
2014-01-02 14:40
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题目:113 - Power of Cryptography
题目大意:求k,k^n = p;
解题思路:k = p ^ (1/n);并且double的上限是1.7E+308,p的范围是1.0E + 101.
题目大意:求k,k^n = p;
解题思路:k = p ^ (1/n);并且double的上限是1.7E+308,p的范围是1.0E + 101.
#include<stdio.h> #include<math.h> int main() { double p, n; while (scanf("%lf%lf", &n, &p) != EOF) { printf("%.lf\n", pow(p, 1/n)); } return 0; }
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