Squarefree number
Squarefree numberTime Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1125 Accepted Submission(s): 274Problem Description
In mathematics, a squarefree number is one which is divisible by no perfect squares, except 1. For example, 10 is square-free but 18 is not, as it is divisible by 9 = 3^2. Now you need to determine whether an integer is squarefree or not.
Input
The first line contains an integer T indicating the number of test cases. For each test case, there is a single line contains an integer N.
Technical Specification 1. 1 <= T <= 20 2. 2 <= N <= 10^18 Output
For each test case, output the case number first. Then output "Yes" if N is squarefree, "No" otherwise.
Sample Input
2 30 75
Sample Output
Case 1: Yes Case 2: No
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题意很简单,就是然你判断一个数能不能被一个数的平方整除(注意这个数的范围是1~10^18)
以下转载于:
http://hi.baidu.com/304467594/blog/item/f02407ea9eeecac62f2e2136.html
/* 要判断给定的n(2<=n<=10^18)是不是squarefree number; 因为给定的数最大达到10^18,所以直接暴力肯定是杯具的TLE; 10^18=10^6^2*10^6; */
下面是神龙的解题思路,我跳过了第三步; /* 1. 计算1000000以下 素数,存于 prime[ ]
2. 计算N的 小于1000000的因子p,如果有平方因子,则No,转 5 否则 N=N/p 3. 消除N的小于 1000000的因子p后,若N=1 , 则Yes,转 5 4. N必为 3种之一: 素数,素数的平方,两个不同素数之积。 因此判断 N是否平方数,是平方数则No,不是则Yes 5. 结束 */
自己的体会:一个数可以写成n个素数的乘积的形式如n=p1^k1*p2^k2*p3^k3....*pn^kn当ki>=2时就是no了;对于pi<10^6所对应的ki是否>=2很容易判断;如果n存在大于10^6的素数的话,最多存在两个这样的素数,当这两个素数相等的时候就no了,其他情况都是yes;
#include<stdio.h>
#include<string.h>
#include<math.h>
#define nmax 1000001
int prime[nmax], plen;
void init() {
memset(prime, -1, sizeof(prime));
int i, j;
for (i = 2; i < nmax; i++) {
if (prime[i]) {
for (j = i + i; j < nmax; j += i) {
prime[j] = 0;
}
}
}
for (i = 2, plen = 0; i < nmax; i++) {
if (prime[i]) {
prime[plen++] = i;
}
}
}
int solve(long long n) {
int i;
for (i = 0; (i < plen) && (prime[i] < n); i++) {
if (n % prime[i] == 0) {
n /= prime[i];
if (n % prime[i] == 0) {
return 0;
}
}
}
return 1;
}
int main() {
init();
int i, t, te, flag;
long long n;
scanf("%d", &t);
for (i = 1; i <= t; i++) {
scanf("%I64d", &n);
flag = solve(n);
printf("Case %d: ", i);
if (flag) {
te = (int) (sqrt(n * 1.0));
if (((long long) (te)) * te == n) {
printf("No\n");
} else {
printf("Yes\n");
}
} else {
printf("No\n");
}
}
return 0;
}