题意:已知n个矩阵(下标从1开始),求下标x~y区间矩阵的乘积。最多m次询问,n ( 1 <= n <= 30,000) and m ( 1 <= m <= 30,000)。
分析:
1、矩阵初始化为单位矩阵,因为要做乘积,E*A=A。
2、因为输出矩阵的所有值范围在0~r-1,所以要对r取余。
#pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) typedef long long ll; typedef unsigned long long llu; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const ll LL_INF = 0x3f3f3f3f3f3f3f3f; const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e4; const double pi = acos(-1.0); const double eps = 1e-8; const int MAXN = 30000 + 10; const int MAXT = 10000 + 10; using namespace std; struct Matrix{ int ma[2][2]; Matrix(){ ma[0][0] = ma[1][1] = 1; ma[0][1] = ma[1][0] = 0; } }num[MAXN << 2]; int r; Matrix mul(Matrix a, Matrix b){ Matrix ans; for(int i = 0; i < 2; ++i){ for(int j = 0; j < 2; ++j){ ans.ma[i][j] = 0; for(int k = 0; k < 2; ++k){ (ans.ma[i][j] += (a.ma[i][k] * b.ma[k][j]) % r) %= r; } } } return ans; } void add_up(int id){ num[id] = mul(num[id << 1], num[id << 1 | 1]); } void add(int l, int r, int cur, Matrix x, int id){ if(l == r){ num[id] = x; return; } int mid = (l + r) >> 1; if(cur <= mid){ add(l, mid, cur, x, id << 1); } else add(mid + 1, r, cur, x, id << 1 | 1); add_up(id); } Matrix query(int L, int R, int l, int r, int id){ if(l >= L && r <= R){ return num[id]; } int mid = (l + r) >> 1; Matrix ans; if(L <= mid){ ans = mul(ans, query(L, R, l, mid, id << 1)); } if(R >= mid + 1){ ans = mul(ans, query(L, R, mid + 1, r, id << 1 | 1)); } return ans; } int main(){ int n, m; bool flag = true; while(scanf("%d%d%d", &r, &n, &m) == 3){ if(flag) flag = false; else printf("\n"); for(int i = 1; i <= n; ++i){ Matrix tmp; for(int j = 0; j < 2; ++j){ for(int k = 0; k < 2; ++k){ scanf("%d", &tmp.ma[j][k]); } } add(1, n, i, tmp, 1); } while(m--){ int x, y; scanf("%d%d", &x, &y); Matrix t = query(x, y, 1, n, 1); printf("%d %d\n", t.ma[0][0], t.ma[0][1]); printf("%d %d\n", t.ma[1][0], t.ma[1][1]); if(m) printf("\n"); } } return 0; }