Submission #1203397

#TimeUsernameProblemLanguageResultExecution timeMemory
1203397KavelmydexHexagonal Territory (APIO21_hexagon)C++20
9 / 100
1 ms328 KiB
/* I: 1. The path does not conatin layers = Must compute: 1. the number of cells in the territory. ====================== ====================== 3 1 1 1 1 3 1 5 1 ans: 3 + 2*1 = 5 3 1 2 1 1 3 1 5 1 ans: 7 3 10000 0 4 998 2 999 6 999 ans: 1001*(1000)/2 * 10000 = 5,005,000,000 = 4999965 */ #include "hexagon.h" #include <bits/stdc++.h> using namespace std; #define ll long long #define vi vector<int> #define pi pair<int,int> #define pb push_back #define sz(x) ((int)x.size()) #define sp ' ' #define endl "\n" #define all(x) (x).begin(),(x).end() #define rep(i,x,n) for(int i=x; i<=n; ++i) #define For(i,n) rep(i,0,n-1) #define ff first #define ss second #define ld long double #define mp make_pair const ll mxN=6e5+10,OO=2e9,mod=1e9+7; ll exp(ll b,ll p){ ll res=1; while(p>0){ if(p&1)res=res*b%mod; b=b*b%mod; p/=2; } return res; } const ll inv=exp(6,mod-2); const int dx[]{0,0,-1,1}, dy[]{1,-1,0,0}; void cmn(int &a,int b){a = min(a,b);} void cmx(int &a,int b){a = max(a,b);} int draw_territory(int N, int A, int B,vi D, vi L){ sort(all(D)); ll v=max(L[0],L[1]),ans=0,c=v+1, a=A,b=B; if(N==3){ assert((D==((vi){1,3,5}) || D==((vi){2,4,6}))); assert(L[2]<=v); } c = c * (c+1)/2; ans = (c%mod) * a % mod; if(B){ // c= v*(v+1)*(2*v+1)/6; c = (v*(v+1)%mod) * (2*v+1) % mod * inv % mod; c += v*(v+1)/2; c%=mod; ans += c*b % mod; ans %=mod; } return ans; }
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