HDU 5104 - Primes Problem (枚举)

题意

找出n1 + n2 + n3 = n && n1 and n2 and n3为素数的三元组数

思路

先打表出1W以内的素数，然后枚举n2、n3。判断n - n2 - n3是否为素数，如果是的话判断该数是否 <= n2。是的话ans + 1

一开始用lower_bound竟然T了？

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-8;
const int MAXN = 10000 + 10;
const int MOD = 1000007;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };  //0123£¬ÉÏÏÂ×óÓÒ
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
vector<int> pri;
int n, vis[MAXN];
void GetPrime()
{
    vis[1] = 1;
    for (int i = 2; i <= MAXN; i++)
        if (!vis[i])
        {
            pri.PB(i);
            for (int j = i * 2; j <= MAXN; j += i) vis[j] = 1;
        }
}
int main()
{
    //ROP;
    ios::sync_with_stdio(0);
    int i, j;
    GetPrime();
    while (cin >> n)
    {
        int ans = 0;
        for (int i = 0; pri[i] < n && i < SZ(pri); i++)
            for (j = i; pri[j] + pri[i] < n && j < SZ(pri); j++)
            {
                if (vis[n - pri[i] - pri[j]] == 0 && n - pri[i] - pri[j] <= pri[i])
                    ans++;
            }
        cout << ans << endl;
    }
    return 0;
}