"BZOJ3505" [CQOI2014] triangular numbers
Do not directly ask this question, consider the inclusion and exclusion, first select the three points regardless of whether legitimate program number $ C _ {(n + 1) * (m + 1)} ^ {3} $, then Save to the three-point line number just fine. Obviously can not be enumerated endpoint, we can enumerate consider two points x, y difference i, j, then the number of intermediate points of the whole (gcd (i, j) -1), so that a plurality of square, so (n-i + 1) * (m-j + 1) * (gcd (i, j) -1) * 2, is multiplied by 2 because there are two diagonal lines, but when i = 0 or j = 0 it is you can not take the 2.
1 #include<iostream> 2 #include<cstdio> 3 #define int LL 4 #define LL long long 5 using namespace std; 6 int n,m; 7 int gcd(int a,int b){return !b?a:gcd(b,a%b);} 8 LL ans; 9 signed main() 10 { 11 cin>>n>>m;n++,m++; 12 ans=((n*m)*(n*m-1)*(n*m-2)/3)/2; 13 n--,m--; 14 for(int i=0;i<=n;i++) 15 for(int j=0;j<=m;j++) 16 if(i||j) 17 if(!i||!j)ans-=(n-i+1)*(m-j+1)*(gcd(i,j)-1); 18 else ans-=(n-i+1)*(m-j+1)*(gcd(i,j)-1)*2; 19 cout<<ans<<endl; 20 }