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题目链接:http://acm.hust.edu.cn/vjudge/problem/36250
题意:求那个式子。
设3k+7=x,则化简成 Sn=Σ(k=1~n) (((x-1)!+1/x)-[(x-1)!/x])
根据威尔逊定理,假如一个数p是素数,则这个数满足:(p-1)!=-1 (mod p)即 (p-1)!-1=0(mod p)。
由于被减数满足此条件,而减数表示向下取整。则被减数整除,减数一定是向下取整的。所以结果减数比被减数要小1,否则减数和被减数相等,即为0。问题转换成了求3k+7是否是素数,打表即可。
1 /* 2 ━━━━━┒ギリギリ♂ eye! 3 ┓┏┓┏┓┃キリキリ♂ mind! 4 ┛┗┛┗┛┃\○/ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &(a)) 43 #define Rll(a) scanf("%lld", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onenum(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long Uint; 62 typedef pair<int, int> pii; 63 typedef pair<LL, LL> pLL; 64 typedef pair<string, LL> psi; 65 typedef map<string, LL> msi; 66 typedef vector<LL> vi; 67 typedef vector<LL> vl; 68 typedef vector<vl> vvl; 69 typedef vector<bool> vb; 70 71 const int maxn = 6000100; 72 bool isprime[maxn]; 73 int n; 74 int dp[maxn/6]; 75 76 void init() { 77 int nn = int(sqrt(maxn)); 78 memset(isprime, true, sizeof(isprime)); 79 isprime[0] = isprime[1] = 0; 80 for(int i = 2; i <= nn; i++) { 81 if(isprime[i]) { 82 int p = maxn / i; 83 for(int j = 2; j <= p; j++) { 84 isprime[j*i] = 0; 85 } 86 } 87 } 88 } 89 90 int main() { 91 // FRead(); 92 init(); 93 Cls(dp); dp[1] = 0; 94 For(i, 2, 1000010) { 95 dp[i] = dp[i-1]; 96 if(isprime[3*i+7]) dp[i]++; 97 } 98 int T; 99 Rint(T); 100 W(T) { 101 Rint(n); 102 cout << dp[n] << endl; 103 } 104 RT 0; 105 }
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原文地址:http://www.cnblogs.com/vincentX/p/5824106.html