LightOJ

题意：

给定n，找到两个素数，想加得到n，问有多少中方法；

思路：

筛法打个素数表，枚举小于等于（n/2）的所有素数prim[ i ]，判（n - prim[ i ]）是不是素数

#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<set>#include<queue>#include<stack>#include<map>#define PI acos(-1.0)#define in freopen("in.txt", "r", stdin)#define out freopen("out.txt", "w", stdout)using namespace std;typedef long long ll;typedef unsigned long long ull;const int maxn = 1e7 + 7, maxd = 670000 + 7, mod = 1e9 + 7;const int INF = 0x7f7f7f7f;int T, n, cnt;bool p[maxn];vector<int> prim;void init() {    memset(p, 1, sizeof p);    for(int i = 4; i < maxn; i += 2) p[i] = 0;    cnt = 0;    prim.push_back(2);    for(int i = 3; i < maxn; i += 2) {        if(p[i]) prim.push_back(i);        for(int j = 0; j < prim.size() && i*prim[j] < maxn; ++j)            p[i*prim[j]] = 0;    }}void solve() {    int ans = 0;    for(int i = 0; i < prim.size() && 2*prim[i] <= n; ++i) {        if(p[n-prim[i]]) ans++;    }    cout << ans << endl;}int main() {    init();    scanf("%d", &T);    for(int tt = 1; tt <= T; ++tt) {        scanf("%d", &n);        printf("Case %d: ", tt);        solve();    }    return 0;}