TrickGCD

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 549    Accepted Submission(s): 211


Problem Description You are given an array A , and Zhu wants to know there are how many different array B satisfy the following conditions?

* 1≤Bi≤Ai
* For each pair( l , r ) (1

≤l≤r≤n) , gcd(bl,bl+1...br)≥2
Input The first line is an integer T(1≤T≤10) describe the number of test cases.

Each test case begins with an integer number n describe the size of array A.

Then a line contains n numbers describe each element of A

You can assume that 1≤n,Ai≤105
Output For the kth test case , first output "Case #k: " , then output an integer as answer in a single line . because the answer may be large , so you are only need to output answer m

od 109+7
Sample Input 1 4 4 4 4 4
Sample Output Case #1: 17

給你序列a，讓你構造序列b，要求 1<=b[i]<=a[i],且b序列的gcd>=2。問你方案數。

容易想到的就是我們列舉整個序列的gcd，然後a[i]/gcd就是i位置能夠填的數的個數，然後每個位置累積就能得到數列為gcd時的方案數。

最後容斥一下累加就是答案。但是最大gcd可以是100000和明顯這樣做n2，會超時。

那麼我們把a[i]/gcd的放在一起，然後用快速冪直接求出值。

#include <bits/stdc++.h>

using namespace std;
const int MAXN = 1e5+7;
const long long mod = 1e9+7;
typedef long long LL;
int n;
int a[MAXN];//a陣列統計每個數的個數
int sum[MAXN];//字首和
LL num[MAXN];//num[i]表示數列gcd為i時的的累積
LL dp[MAXN];//dp[i]表示容斥完之後也就是最終結果 gcd為i時的累積
LL quick_pow(int a,int b)
{
    long long sum = 1,base = a;
    while(b)
    {
        if(b & 1)sum = sum*base%mod;
        base  = base*base%mod;
        b >>= 1;
    }
    return sum;
}

int main()
{
    int t;
    scanf("%d",&t);
    int ca = 0;
    while(t--)
    {
        int x;
        scanf("%d",&n);
        memset(a,0,sizeof a);
        for(int i = 0; i < n; ++i)
        {
            scanf("%d",&x);
            a[x]++;
        }
        memset(sum,0,sizeof sum);
        for(int i = 1; i <= 100000; ++i)sum[i] = sum[i-1] + a[i];
        int flag = 1;
        for(int i = 2; i <= 100000; ++i)//列舉gcd
        if(!flag)num[i] = 0;
        else
        {
            num[i] = 1;
            for(int j = 0; j <= 100000; j+=i)
            {
                int b = sum[min(j+i-1,100000)] - sum[max(j-1,0)];//注意邊界
                int a = j/i;
                if(!a && b)//如果存在比當前i小的數，那麼後面的肯定都不符合
                {
                    num[i] = 0;
                    flag = 0;
                    break;
                }
                num[i] = num[i]*quick_pow(a,b)%mod;
            }
        }
        for(int i = 100000; i >= 2; --i)
        {
            dp[i] = num[i];
            for(int j = i<<1; j <= 100000; j +=i)
            {
                dp[i] -= dp[j];
                dp[i] = (dp[i]%mod+mod)%mod;
            }
        }
        LL ans = 0;
        for(int i = 2; i <= 100000; ++i)ans = (ans+dp[i])%mod;
        printf("Case #%d: %I64d\n",++ca,ans);
    }
    return 0;
}