[数论] ACM 2015 Changchun B Count a*b

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题意：

题解：

#include<cstdio>#include<cstring>#include<algorithm>#include<set>#include<map>#include<vector>#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)#define per(i,j,k) for(int i=(int)j;i>=(int)k;i--)using namespace std;typedef long long LL;typedef double db;int n,P;inline int S1(int x){return (x*1ll*(x+1)/2)%P;}inline int S2(int x){int x1=x;int x2=x+1;int x3=2*x+1;if(x1%2==0)x1/=2;else x2/=2;if(x1%3==0)x1/=3;else if(x2%3==0)x2/=3;else x3/=3;return (x1*1ll*x2%P)*1ll*x3%P;}void Main(){int ans=0;int ans2=0;int pre1,pre2;pre1=pre2=0;for(int i=1;i<=n;i++){int j=n/(n/i);int ss2=S2(j);int ss1=S1(j);ans2=(ans2+(ss2+P-pre2)*1ll*(n/i))%P;ans=(ans+S1(n/i)*1ll*(ss1+P-pre1))%P;pre1=ss1;pre2=ss2;i=j;}printf("%d\n",(ans2+P-ans)%P);}int main(){freopen("phy.in","r",stdin);freopen("phy.out","w",stdout);int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&P);Main();}return 0;}

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