
题面
如果是一条链的话,我们直接搞个转移矩阵,矩阵乘法即可
破环为链,并硬点一个端点黄色/绿色,然后讨论一下即可
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using namespace std; typedef long long ll; const ll mod = 1000000007; struct { ll a[2][2]; friend inline matrix operator * (matrix a, matrix b) { matrix ans; memset(&ans, 0, sizeof ans); for(int i = 0; i < 2; i++) for(int k = 0; k < 2; k++) for(int j = 0; j < 2; j++) ans[i][j] = (ans[i][j] + 1ll * a[i][k] * b[k][j]) % mod; return ans; } ll* operator [] (ll x) {return a[x];} } res, ans; int main() { int t; long long n; scanf("%d", &t); while(t--) { scanf("%lld", &n); ans[0][0] = ans[1][1] = 1; ans[0][1] = ans[1][0] = 0; res[0][0] = res[0][1] = res[1][0] = 1; res[1][1] = 0; for(n--; n; n >>= 1, res = res * res) if(n & 1) ans = ans * res; printf("%lldn", ((ans[0][0] + ans[0][1]) % mod + ans[1][0]) % mod); } return 0; }
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