UVA 10622 - Perfect P-th Powers(数论)
2014-06-22 20:05
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UVA 10622 - Perfect P-th Powers
题目链接题意:求n转化为b^p最大的p值
思路:对n分解质因子,然后取所有质因子个数的gcd就是答案,但是这题有个坑啊,就是输入的可以是负数,负数的情况比较特殊,p只能为奇数,这时候是要把答案不断除2除到为奇数即可。
代码:
#include <stdio.h>
#include <string.h>
#include <math.h>
long long n;
int prime[333333], vis[333333], m = 0;
int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
int solve() {
long long nn = n;
if (n < 0) n = -n;
int ans = 0;
for (int i = 0; i < m && prime[i] <= n; i++) {
int count = 0;
while (n % prime[i] == 0) {
count++;
n /= prime[i];
}
ans = gcd(ans, count);
}
if (ans == 0) ans = 1;
if (nn < 0) {
while (ans % 2 == 0) {
ans /= 2;
}
}
return ans;
}
int main() {
for (int i = 2; i < 333333; i++) {
if (vis[i]) continue;
prime[m++] = i;
for (int j = i; j < 333333; j += i) {
vis[j] = 1;
}
}
while (~scanf("%lld", &n) && n) {
printf("%d\n", solve());
}
return 0;
}
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