Submission #609821

#TimeUsernameProblemLanguageResultExecution timeMemory
609821dualityMutating DNA (IOI21_dna)C++17
0 / 100
32 ms6928 KiB
#define DEBUG 0 #include <bits/stdc++.h> using namespace std; #if DEBUG // basic debugging macros int __i__,__j__; #define printLine(l) for(__i__=0;__i__<l;__i__++){cout<<"-";}cout<<endl #define printLine2(l,c) for(__i__=0;__i__<l;__i__++){cout<<c;}cout<<endl #define printVar(n) cout<<#n<<": "<<n<<endl #define printArr(a,l) cout<<#a<<": ";for(__i__=0;__i__<l;__i__++){cout<<a[__i__]<<" ";}cout<<endl #define print2dArr(a,r,c) cout<<#a<<":\n";for(__i__=0;__i__<r;__i__++){for(__j__=0;__j__<c;__j__++){cout<<a[__i__][__j__]<<" ";}cout<<endl;} #define print2dArr2(a,r,c,l) cout<<#a<<":\n";for(__i__=0;__i__<r;__i__++){for(__j__=0;__j__<c;__j__++){cout<<setw(l)<<setfill(' ')<<a[__i__][__j__]<<" ";}cout<<endl;} // advanced debugging class // debug 1,2,'A',"test"; class _Debug { public: template<typename T> _Debug& operator,(T val) { cout << val << endl; return *this; } }; #define debug _Debug(), #else #define printLine(l) #define printLine2(l,c) #define printVar(n) #define printArr(a,l) #define print2dArr(a,r,c) #define print2dArr2(a,r,c,l) #define debug #endif // define #define MAX_VAL 999999999 #define MAX_VAL_2 999999999999999999LL #define EPS 1e-6 #define mp make_pair #define pb push_back // typedef typedef unsigned int UI; typedef long long int LLI; typedef unsigned long long int ULLI; typedef unsigned short int US; typedef pair<int,int> pii; typedef pair<LLI,LLI> plli; typedef vector<int> vi; typedef vector<LLI> vlli; typedef vector<pii> vpii; typedef vector<plli> vplli; // ---------- END OF TEMPLATE ---------- #include "dna.h" int pre[100001][3][3],num[3][3]; int pos(char c) { if (c == 'A') return 0; else if (c == 'C') return 1; else return 2; } void init(string a,string b) { int i,n = a.size(); for (i = 0; i < n; i++) { memcpy(pre[i+1],pre[i],sizeof(pre[i])); pre[i+1][pos(a[i])][pos(b[i])]++; } } int get_distance(int x,int y) { int i,j; for (i = 0; i < 3; i++) { for (j = 0; j < 3; j++) num[i][j] = pre[y+1][i][j]-pre[x][i][j]; } int ans = 0; for (i = 0; i < 3; i++) { j = (i+1) % 3; int m = min(num[i][j],num[j][i]); ans += m,num[i][j] -= m,num[j][i] -= m; } if ((num[0][1] != num[1][2]) || (num[1][2] != num[2][0])) return -1; else ans += 2*num[0][1]; if ((num[1][0] != num[2][1]) || (num[2][1] != num[1][0])) return -1; else ans += 2*num[1][0]; return ans; }
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