题目描述
We define Shuaishuai-Number as a number which is the sum of a prime square(平方), prime cube(立方), and prime fourth power(四次方).
The first four Shuaishuai numbers are:
How many Shuaishuai numbers in [1,n]? (1<=n<=50 000 000)
输入描述:
The input will consist of a integer n.
输出描述:
You should output how many Shuaishuai numbers in [1...n]
示例1
输入
28
输出
1
说明
There is only one Shuaishuai number
题解
暴力打标。
把所有满足要求的数组都存进数组，排序后去重，每次询问二分即可。
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 50000000;
int a[1200000 + 10];
int sz = 0, cnt = 0;
int b[1200000 + 10];
bool noprime[maxn + 10];
int n;
void init() {
noprime[1] = 1;
for(int i = 2; i <= maxn; i ++) {
if(noprime[i]) continue;
for(int j = i + i; j <= maxn; j = j + i) {
noprime[j] = 1;
}
}
for(int i = 1; i * i <= maxn; i ++) {
if(noprime[i]) continue;
for(int j = 1; i * i + j * j * j <= maxn; j ++) {
if(noprime[j]) continue;
for(int k = 1; i * i + j * j * j + k * k * k * k <= maxn; k ++) {
if(noprime[k]) continue;
a[sz ++] = i * i + j * j * j + k * k * k * k;
}
}
}
sort(a, a + sz);
b[cnt ++] = a[0];
for(int i = 1; i < sz; i ++) {
if(a[i] == a[i - 1]) continue;
b[cnt ++] = a[i];
}
}
int main() {
init();
while(~scanf("%d", &n)) {
int L = 0, R = cnt - 1, pos = -1;
while(L <= R) {
int mid = (L + R) / 2;
if(b[mid] <= n) pos = mid, L = mid + 1;
else R = mid - 1;
}
printf("%d\n", pos + 1);
}
return 0;
}
