「The 15th Zhejiang Provincial」F.Now Loading!!!(数论+二分)

描述

传送门：我是传送门

DreamGrid has nn integers a1,a2,…,ana1,a2,…,an. DreamGrid also has mm queries, and each time he would like to know the value of

∑1≤i≤n⌊ai⌈logpai⌉⌋∑1≤i≤n⌊ai⌈logp⁡ai⌉⌋

for a given number pp, where ⌊x⌋=maxy∈ℤ∣y≤x⌊x⌋=maxy∈Z∣y≤x, ⌈x⌉=miny∈ℤ∣y≥x⌈x⌉=miny∈Z∣y≥x.

输入

There are multiple test cases. The first line of input is an integer TT indicating the number of test cases. For each test case:

The first line contains two integers nn and mm (1≤n,m≤5×1051≤n,m≤5×105) — the number of integers and the number of queries.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (2≤ai≤1092≤ai≤109).

The third line contains mm integers p1,p2,…,pmp1,p2,…,pm (2≤pi≤1092≤pi≤109).

It is guaranteed that neither the sum of all nn nor the sum of all mm exceeds 2×1062×106.

输出

For each test case, output an integer (∑i=1mi⋅zi)mod109(∑i=1mi⋅zi)mod109, where zizi is the answer for the ii-th query.

样例

输入

2
3 2
100 1000 10000
100 10
4 5
2323 223 12312 3
1232 324 2 3 5

输出

11366
45619

思路

这道题的所有题解我全都看了一遍，基本思路大致搞明白，但是卡在最后的一句代码上

1	ans = (ans+tmp*j)%mod;

最后我发现(∗j∗j)是题目要求的，是时候换个眼睛脑子了😤

刚开始毫无头绪(现在看完别人的题解也是没有头绪···

现场赛之前再回头看一下这个题目吧，这次不写题解了，我现在也实在写不出了来几句···

另外，为什么ZOJ爆内存提示段错误？？？？

代码

/*
 * =================================================================
 *
 *       Filename:  zoj4029.cpp
 *
 *           Link:  http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4029
 *
 *        Version:  1.0
 *        Created:  2018/10/22 21时11分39秒
 *       Revision:  none
 *       Compiler:  g++
 *
 *         Author:  杜宁元 (https://duny31030.top/), duny31030@126.com
 *   Organization:  QLU_浪在ACM
 *
 * =================================================================
 */
#include <bits/stdc++.h>
using namespace std;
#define clr(a, x) memset(a, x, sizeof(a))
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define pre(i,a,n) for(int i=n;i>=a;i--)
#define ll long long
#define max3(a,b,c) fmax(a,fmax(b,c))
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int mod = 1e9;
const int N = 5e5+100;
ll a[N];
ll t,n,m,x;
ll sum[N][32];

// 1.对于每一次询问p，枚举分母 i 时，可以找出a中分母等于 i 的那一段
// 预处理前缀和，用于此时直接加。

int main()
{
    ios
#ifdef ONLINE_JUDGE
#else
        freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
#endif
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld %lld",&n,&m);
        rep(i,1,n)  scanf("%lld",&a[i]);
        // rep(i,1,m)  scanf("%d",&p[i]);
        sort(a+1,a+1+n);
        // rep(i,1,n)  printf("%lld ",a[i]);   printf("\n");
        // 打表预处理
        for(ll k = 1;k <= 30;k++)
            for(ll i = 1;i <= n;i++)
                sum[i][k] = sum[i-1][k]+a[i]/k;

        ll ans = 0;
        for(ll j = 1;j <= m;j++)
        {
            scanf("%lld",&x);
            ll pos;
            ll up = x;
            ll k = 1;
            ll tmp = 0;
            for(ll i = 1;i <= n;i = pos+1)
            {
                // printf("up = %lld\n",up);
                pos = i-1;
                ll l = i,r = n;
                while(l <= r)
                {
                    ll mid = (l+r)>>1;
                    if(a[mid] <= up)
                    {
                        pos = mid;
                        l = mid+1;
                    }
                    else
                        r = mid-1;
                }
                tmp = (tmp+sum[pos][k]-sum[i-1][k]+mod)%mod;
                k++;
                up *= x;
            }
            //// printf("tmp = %lld j = %lld tmp*j = %lld ans = %lld\n",tmp,j,tmp*j,ans+tmp*j);
            ans = (ans+tmp*j)%mod;
        }
        printf("%lld\n",ans);
    }

    fclose(stdin);
    // fclose(stdout);
    return 0;
}