題目描述

We define Shuaishuai-Number as a number which isthe 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)

輸入描述:

The input will consist of a integer n.

輸出描述:

You should output how many Shuaishuai numbers in [1...n]

示例1

輸入

28

輸出

1

說明

There is only one Shuaishuai number

題解

暴力打標。

把所有滿足要求的數組都存進數組，排序后去重，每次詢問二分即可。

#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
", pos + 1);
  }
  return 0;
}

　　

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