Codeforces 396B On Sum of Fractions 数论

题目链接：Codeforces 396B On Sum of Fractions

题解来自：http://blog.csdn.net/keshuai19940722/article/details/20076297

题目大意：给出一个n，ans = ∑（2≤i≤n）1/(v(i)*u(i)), v(i)为不大于i的最大素数，u(i)为大于i的最小素数， 求ans，输出以分式形式。

解题思路：一開始看到这道题1e9，暴力是不可能了，没什么思路，后来在纸上列了几项，突然想到高中时候求等差数列时候用到的方法。详细叫什么不记得了。

1/(2*3) = (1/2  - 1/3) * 1/(3-2);

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

然后值为1/(2*3)的个数有（3-2）个，为1/(3*5)的有（5-3）个。

这样如果有n，v = v(n),  u = u(n)； 

1/(2*3) + 1/(3*5) * (5-3) + ...... + 1/(v*u) * (n-v+1)  (注意最后不是u-v个）

= 1/2 - 1/3 + 1/3 - 1/5 + ........ -1/v + 1/(v*u) *(n-v+1)

= 1/2 - 1/v + 1/(v*u)*(n-v+1)

p = u*v + 2*(n-v-u+1); q = 2*u*v;

记得约分，然后u和v就用枚举的方式。

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代码

#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
#define 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;
}