UVA:11426 GCD - Extreme (II)

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题目地址：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2421

本题跟欧拉函数有关！

设f[n]=gcd(1,n)+gcd(2,n)+....+gcd(n-1,n)

如何快速求f[n]呢？

i=gcd(x,n),

g(n,i)为gcd(x,n)==i的个数，则

f[n]=sum(i*g(n,i))(i为n的约数)

如何求g(n,i)呢？

则

gcd(x,n)=i,

gcd(x/i,n/i)=1,

所以g(n,i)=phi(n/i)

本题需要快速求phi数组，否则会超时!

上代码(注意本题在C++11 4.8.2 - GNU C++编译通过)：

#include<cstdio>#include<cstring>const int maxn = 4000000;typedef long long LL;LL S[maxn + 1], f[maxn + 1];LL phi[maxn+1];bool flag[maxn+1];LL prim[maxn+1];void phi_table(){    int i;    int len = 0;    phi[1] = 0;    memset(flag, false, sizeof(flag));    for (i = 2; i <= maxn; i++){        if (!flag[i]){            phi[i] = i - 1;            prim[len++] = i;        }        for (int j =0; j < len&&prim[j] * i <= maxn; j++){            flag[i*prim[j]] = true;            if (i%prim[j] == 0){                phi[i*prim[j]] = phi[i] * prim[j];                break;            }            else{                phi[i*prim[j]] = phi[i] * (prim[j] - 1);            }        }    }}int main(){    int i;    int n;    phi_table();    memset(f, 0, sizeof(f));    for (i = 1; i <= maxn; i++){        for (n = i * 2; n <= maxn; n += i) f[n] += i*phi[n / i];    }    S[2] = f[2];    for (n = 3; n <= maxn; n++) S[n] = S[n - 1] + f[n];    while (scanf("%d", &n) == 1 && n){        printf("%lld\n", S[n]);    }    return 0;}

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