HDU6053 TrickGCD（2017多校第2场）

TrickGCD

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1218    Accepted Submission(s): 464

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 mod 109+7

 

Sample Input

144 4 4 4

 

Sample Output

Case #1: 17

 

Source

2017 Multi-University Training Contest - Team 2

 

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题意：给你n个数字，每个位置的数字可以小于等于a[i]，求所有gcd(l,r)都满足大于等于2的情况数

解题思路：枚举gcd的情况，每种gcd的情况等于所有a[i]/gcd的乘积，但这显然会超，所以需要优化，我们可以分块处理，如枚举5时，5 6 7 8 9对应的都是1个，所以我们可以按gcd分块，而且越大块越少，快内用快速幂加速。 得到一个dp数组dp[i]表示gcd为i的方案数，最后容斥搞一搞

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <string>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define LL long longconst LL mod=1e9+7;const int INF=0x3f3f3f3f;#define MAXN 100005int pre[MAXN];LL a[MAXN];LL dp[MAXN];LL qpow(LL a,LL b){    LL ans=1;    while(b)    {        if(b&1) ans=(ans*a)%mod;        b>>=1,a=(a*a)%mod;    }    return ans;}int main(){    int T,n;    int q=1;    for(scanf("%d",&T); T--;)    {        scanf("%d",&n);        memset(pre,0,sizeof pre);        for(int i=0; i<n; i++)        {            scanf("%lld",&a[i]);            pre[a[i]]++;        }        for(int i=1; i<MAXN; i++) pre[i]+=pre[i-1];        for(int i=2; i<MAXN; i++)        {            dp[i]=1LL;            for(int j=0; j<MAXN; j+=i)            {                int cnt;                if (j == 0) cnt = pre[j + i - 1];                else if (j + i - 1 > 100000) cnt = pre[100001] - pre[j - 1];                else  cnt=pre[j+i-1]-pre[j-1];                if(j/i==0&&cnt) {dp[i]=0;break;}                dp[i]=(dp[i]*qpow(j/i,(LL)cnt))%mod;            }        }        LL ans=0;        for(int i=a[n-1]; i>1; i--)        {            for(int j=i+i; j<=a[n-1]; j+=i)            {                dp[i]-=dp[j];                dp[i]=(dp[i]%mod+mod)%mod;            }            ans+=dp[i];            ans%=mod;        }        printf("Case #%d: %lld\n",q++,ans);    }    return 0;}