LightOj 1006 - Hex-a-bonacci
2014-09-04 12:47
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#include <iostream> #include <cstdio> #include <string> #include <cstring> #include <algorithm> #include <cmath> #include <vector> #include <list> #include <deque> #include <queue> #include <cctype> #include <map> #include <set> #include <bitset> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iomanip> #include <cstdlib> #include <ctime> #include <cassert> #include <limits> #include <fstream> using namespace std; #define mem(A, X) memset(A, X, sizeof A) #define pb(x) push_back(x) #define mp(x,y) make_pair((x),(y)) #define vi vector<int> #define all(x) x.begin(), x.end() #define foreach(e,x) for(__typeof(x.begin()) e=x.begin();e!=x.end();++e) #define sz(x) (int)((x).size()) #define sl(a) strlen(a) #define rep(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i)) #define Rep(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i)) #define min3(a,b,c) min(a,min(b,c)) #define max3(a,b,c) max(a,max(b,c)) #define dbg(a) cout << a << endl; #define fi first #define se second typedef long long int64; inline int nextInt() { int x; scanf("%d", &x); return x; } inline int64 nextI64() { int64 d; cin >> d; return d; } inline char nextChr() { scanf(" "); return getchar(); } inline string nextStr() { string s; cin >> s; return s; } inline double nextDbf() { double x; scanf("%lf", &x); return x; } inline int64 nextlld() { int64 d; scanf("%lld", &d); return d; } inline int64 next64d() { int64 d; scanf("%I64d", &d); return d; } int gcd(const int64 &a, const int64 &b) { return b == 0 ? a : gcd(b, a % b); } int64 int64pow(int64 a, int64 b){ if (b == 0) return 1; int64 t = int64pow(a, b / 2); if (b % 2) return t * t * a; return t * t; } const int inf = 1 << 30; const double eps = 1e-8; const double pi = acos(-1.0); const int MAX_N = 10005; const int MOD = 10000007; int cas = 0, T; int64 a, b, c, d, e, f, n; int64 arr[MAX_N]; void solve() { arr[0] = a; arr[1] = b; arr[2] = c; arr[3] = d; arr[4] = e; arr[5] = f; Rep(i, 6, 10000) { arr[i] = (arr[i - 1] + arr[i - 2] + arr[i - 3] + arr[i - 4] + arr[i - 5] + arr[i - 6]) % MOD; } cout << "Case " << ++cas << ": " << arr % MOD << endl; } int main() { cin >> T; while (cin >> a >> b >> c >> d >> e >> f >> n) { solve(); } return 0; }
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