题意：
题解：
#include #include #include #include #include #include #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; }
![[数论] ACM 2015 Changchun B Count a*b_第1张图片](http://img.e-com-net.com/image/info8/2e0a08092ab34b5bafa2a5a7c5827d75.jpg)
![[数论] ACM 2015 Changchun B Count a*b_第2张图片](http://img.e-com-net.com/image/info8/fc2ce75872664b4ebdc040ec1be0c6c3.jpg)
![[数论] ACM 2015 Changchun B Count a*b_第3张图片](http://img.e-com-net.com/image/info8/961ea4aa7e3e4a26a1c32fbe30e0529f.jpg)
![[数论] ACM 2015 Changchun B Count a*b_第4张图片](http://img.e-com-net.com/image/info8/bd97a8ef86a94f279d145d33a98bf5b8.jpg)
