Title Link: codeforces 396B on Sum of fractions

From: http://blog.csdn.net/keshuai19940722/article/details/20076297

Title: Give a N,ans =∑ (2≤i≤n) 1/(V (i) *u (i)), V (i) is not greater than the maximum number of primes, u (i) is greater than the smallest prime, for ans, the output in fractional form.

Problem-solving ideas: A beginning to see this problem 1e9, violence is impossible, no idea, and later on the paper on a few items, suddenly thought of high school when asked arithmetic progression time to use the method. I don't remember the details.

1/(2*3) = (1/2-1/3) * 1/(3-2);

1/(3*5) = (1/3-1/5) * 1/(5-3);

Then the number of values is 1/(2*3) has (3-2), 1/(3*5) has (5-3).

So if there is n,v = V (n), u = u (n);

1/(2*3) + 1/(3*5) * (5-3) + ... + 1/(v*u) * (n-v+1) (note that the last one is not u-v)

= 1/3 + 1/3-1/5 + ... -1/v + 1/(v*u) * (n-v+1)

= 1/v + 1/(v*u) * (n-v+1)

p = u*v + (n-v-u+1); Q = 2*u*v;

Remember numerator, and then you and V are enumerated in the way.

Learn to:

1, to find a very large number of n nearest to his prime, be able to sift out the prime in sqrt (n), and then, the inferred prime is not an approximate number of n---in fact, with the help of a pair of approximate, small always <=sqrt (n)complexity of time. <SQRT (N) O (1), >SQRT (n), O (n )2, suppose at first glance did not find the law or no train of thought. Simulate several numbers yourself. A continuous simulation. not limited to sample input. Find the rules for yourself .

int Prmcnt;bool is[n];int prm[m];int getprm (int N) { int i,j,k=0; int s,e= (int) (sqrt (0.0+n) +1); CL (is,1); prm[k++]=2;is[0]=is[1]=0; for (i=4;i<n;i+=2) is[i]=0; for (i=3;i<e;i+=2) if (Is[i]) { prm[k++]=i; for (s=i*2,j=i*i; j<n; j+=s) is[j]=0; } for (; i<n;i+=2) if (is[i]) prm[k++]=i; return k;} BOOL judge (int x) { if (x<maxn-1) return is[x]; for (int i=0;i<prmcnt;i++) if (x prm[i] = = 0) return 0; return 1;}

AC Code

#include <cstdio> #include <cstring> #include <algorithm> #include <string> #include < iostream> #include <iomanip> #include <cmath> #include <map> #include <set> #include < queue>using namespace std; #define LS (RT) Rt*2#define RS (RT) Rt*2+1#define ll long long#define ull unsigned long long#de Fine Rep (I,s,e) for (int. i=s;i<e;i++) #define REPE (i,s,e) for (int i=s;i<=e;i++) #define CL (A, B) memset (A,b,sizeof (A )) #define IN (s) freopen (S, "R", stdin) #define OUT (s) freopen (S, "w", stdout) const LL ll_inf = ((ull) ( -1)) >>1;const Double EPS = 1e-8;const int INF = 100000000;const int MAXN = 1e5+5;const int N = maxn;const int m=n;int prmcnt;bool is[n]; int prm[m];int GETPRM (int n) {int i,j,k=0; int s,e= (int) (sqrt (0.0+n) +1); CL (is,1); prm[k++]=2;is[0]=is[1]=0; for (i=4;i<n;i+=2) is[i]=0; for (i=3;i<e;i+=2) if (Is[i]) {prm[k++]=i; for (s=i*2,j=i*i; j<n; j+=s) is[j]=0; } for (; i<n;i+=2) if (is[i]) prm[k++]=i; return k;} BOOL judge (int x) {if (x<maxn-1) return is[x]; for (int i=0;i<prmcnt;i++) if (x prm[i] = = 0) return 0; return 1;} int CaLV (int x) {for (int i=x;i>1;i--) if (judge (i)) return i; }int CalU (int x) {for (int i=x+1;; i++) if (judge (i)) return i;} ll GCD (ll X, ll y) {return y = = 0?X:GCD (y,x%y);} int main () {PRMCNT=GETPRM (MAXN-1); int ncase,n; scanf ("%d", &ncase); while (ncase--) {scanf ("%d", &n); ll v= CaLV (n); ll u= CalU (n); ll up =v*u-2*u-2*v+2*n+2; ll Down=2*v*u; ll TMP=GCD (Up,down); up/=tmp; down/=tmp; cout << up << '/' << down << endl; } return 0;}

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Codeforces 396B on Sum of fractions number theory