codechef Flooring

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Problem Description

Your task is simple.
You need to find the value of
.
As the value could be too large, output it modulo M.
Input

The first contains an integer T, denoting the number of the test cases.
Then there are T lines, each describing a single test case and contains two space separated integers N and M respectively.
Output

For each test case, output the value of summation modulo M on a separate line.

Constraints

1 ≤ M ≤ 100000
There are two types of datasets:
1 ≤ N ≤ 106 , 1 ≤ T ≤ 3000
106 ≤ N ≤ 1010 , 1 ≤ T ≤ 30

Example

Input:

1
4 1000

Output:

373

Explanation

14*4 + 24*2 + 34*1 + 44*1 = 373

题解

数学题折腾了一天……也是醉了。
首先向下取整的操作意味着【N/i】是成阶梯状的。那么我们可以用分块的思想，处理出每一块的和。但是不能暴力求和，这里就涉及到前缀和的思想。
给点提示：1^4+2^4+……+n^4=n(n+1)(2n+1)(3n^2+3n-1)/30，高二上数列好好学……
a mod p= [ (a*m) mod (p*m) ] / m (m为常数)
记得ans要清零，为这个倒是把自家电脑上的问题给解决了。

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#define ll long longusing namespace std;int T;ll n,m,ans;ll calcu(ll x,ll y)//n(n+1)(2n+1)(3n^2+3n-1)/30{ll a,b,c,sum1,sum2,mod;mod=30*m;a=x%mod; b=(a*(a+1))%mod*(2*a+1)%mod; c=(3*a*(a+1)+mod-1)%mod;sum1=(b*c%mod)/30;a=y%mod; b=(a*(a+1))%mod*(2*a+1)%mod; c=(3*a*(a+1)+mod-1)%mod;sum2=(b*c%mod)/30;if(sum2-sum1<0) return sum2-sum1+m;return sum2-sum1;}void work(){ll i,j;ans=0;for(i=1;i<=n;i=j+1)   {j=n/(n/i);    ans=(ans+calcu(i-1,j)*(n/i)%m)%m;   }printf("%lld\n",ans);}int main(){scanf("%d",&T);while(T--)   {scanf("%lld%lld",&n,&m);    work();   }return 0;}

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