湖南大学ACM程序设计新生杯大赛（同步赛） D -Number 【打表+暴力】

时间限制：C/C++ 2秒，其他语言4秒

空间限制：C/C++ 65536K，其他语言131072K

64bit IO Format: %lld

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

分析：素数筛打表，暴力。

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#define ll long long int
using namespace std;
const int maxn = 2e6 + 10;
int a[maxn];
int vis[maxn];
int c[maxn];
int n;
int len;
void get_prime() {
len = 0;
for (int i = 2; i < maxn; i++)
{
if (vis[i])
continue;
a[len++] = i;
for (int j = 2 * i; j < maxn; j += i)
{
vis[j] = 1;
}
}
}
int main()
{
memset(vis, 0, sizeof vis);
memset(c, 0, sizeof c);
scanf("%d", &n);
get_prime();
int ans = 0;
int cs = 0;
for (int i = 0; i < len; i++) {
int tmp1 = a[i] * a[i] * a[i] * a[i];
if (tmp1 > n) break;
for (int j = 0; j < len; j++) {
int tmp2 = a[j] * a[j] * a[j] + tmp1;
if (tmp2 > n) break;
for (int k = 0; k < len; k++) {
int tmp3 =  a[k] * a[k] + tmp2;
if (tmp3 <= n) { c[cs++] = tmp3; }
if (tmp3 > n) break;
}
}
}
sort(c, c + cs);
if (c[0])
ans = 1;
for (int i = 1; i < cs; i++) {
if (c[i] != c[i - 1]) ans++;
}
printf("%d\n", ans);
return 0;
}