DreamGrid has integers . DreamGrid also has queries, and each time he would like to know the value of
for a given number , where , .
Input
There are multiple test cases. The first line of input is an integer indicating the number of test cases. For each test case:
The first line contains two integers and () -- the number of integers and the number of queries.
The second line contains integers ().
The third line contains integers ().
It is guaranteed that neither the sum of all nor the sum of all exceeds .
Output
For each test case, output an integer , where is the answer for the -th query.
Sample Input
2 3 2 100 1000 10000 100 10 4 5 2323 223 12312 3 1232 324 2 3 5
Sample Output
11366 45619
题解： 对于a^n+1与a^(n+1),以a为底取log向上取整后是一样的。 因此可以按区间计算，然后二分查找区间。 注意：先除在求和与先求和在除是不一样的，赛场上没想到 这一点导致这题没过。 代码： #include<bits/stdc++.h> using namespace std; #define ll long long const int maxn=500005; const int mod=1e9; ll a[maxn]; int b[maxn][33]; int main() { int T;scanf("%d",&T); while(T--) { int n,m;scanf("%d%d",&n,&m); ll ma=0,ans=0; for(int i=1;i<=n;i++) { scanf("%lld",&a[i]); ma=max(ma,a[i]); } sort(a+1,a+n+1); for(ll i=1;i<33;i++) { for(int j=1;j<=n;j++) b[j][i]=(b[j-1][i]+a[j]/i)%mod; } for(int i=1;i<=m;i++) { ll x;scanf("%lld",&x); bool bb=0;ll sum=0; int a1=1; for(ll j=1,k=1;;k++) { if(j<=ma)j=j*x; else break; int id=upper_bound(a+a1,a+n+1,j)-a-1; sum=(sum+b[id][k]-b[a1-1][k])%mod; if(sum<0)sum+=mod; a1=id+1; } ans=(ans+i*sum)%mod; } printf("%lld\n",ans); } return 0; }
