hdoj 5212 Code 【数学+思维】

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题目链接：hdoj 5212 Code

Code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 677 Accepted Submission(s): 262

Problem Description
WLD likes playing with codes.One day he is writing a function.Howerver,his computer breaks down because the function is too powerful.He is very sad.Can you help him?

The function:

int calc
{

int res=0;

for(int i=1;i<=n;i++)

for(int j=1;j<=n;j++){  res+=gcd(a[i],a[j])*(gcd(a[i],a[j])-1);  res%=10007;}

return res;

}

Input
There are Multiple Cases.(At MOST 10)

For each case:

The first line contains an integer N(1≤N≤10000).

The next line contains N integers a1,a2,…,aN(1≤ai≤10000).

Output
For each case:

Print an integer,denoting what the function returns.

Sample Input
5
1 3 4 2 4

Sample Output
64

Hint
gcd(x,y) means the greatest common divisor of x and y.

题意：优化程序，快速求解。

思路：我们考虑以d为gcd的数对做出的贡献。
记a[]中d的倍数有cnt[d]个，在d的倍数中，会有若干个数组成数对的gcd为2*d，3*d……。我们设置ans[d]为以d为gcd的数对个数，那么有ans[d] = total - ans[2*d] - ans[3*d] - … 。其中total为cnt[d]*cnt[d]。
发现求解d，需要2*d、3*d……，而求解2*d需要3*d，4*d……。那么我们直接倒着来求就好了。时间复杂度O(nlogn)。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <queue>#include <cmath>#define fi first#define se second#define ll o<<1#define rr o<<1|1#define CLR(a, b) memset(a, (b), sizeof(a))using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MOD = 1e4 + 7;const int MAXN = 1e4 + 10;void add(LL &x, LL y) { x += y; x %= MOD; }LL cnt[MAXN], ans[MAXN];int main(){    int n;    while(scanf("%d", &n) != EOF) {        int Max = 0; CLR(cnt, 0);        for(int i = 1; i <= n; i++) {            int v; scanf("%d", &v);            Max = max(Max, v);            for(int j = 1; j <= sqrt(v); j++) {                if(v % j == 0) {                    cnt[j]++;                    if(j * j != v) {                        cnt[v/j]++;                    }                }            }        }        CLR(ans, 0); LL sum = 0;        for(int i = Max; i >= 1; i--) {            ans[i] = cnt[i] * cnt[i];            for(int j = i * 2; j <= Max; j += i) {                ans[i] -= ans[j];            }            add(sum, ans[i] * (i-1) % MOD * i % MOD);        }        printf("%lld\n", sum);    }    return 0;}

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