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指数循环节,注意a mod p = 0的情况。此时你的循环节如果返回0,这时你会输出1,而实际上应该是0
#include <algorithm> #include <iostream> #include <cstring> #include <cstdio> using namespace std; typedef long long ll; ll n, a, b, c, p; struct Matrix{ ll a[3][3]; void clear(){memset(a, 0, sizeof a);} void set(){clear(); a[0][0] = a[1][1] = a[2][2] = 1;} }mat, ans; Matrix operator * (const Matrix& a, const Matrix& b){ Matrix c; c.clear(); for(int i = 0; i < 3; i ++) for(int j = 0; j < 3; j ++) for(int k = 0; k < 3; k ++) (c.a[i][j] += a.a[i][k] * b.a[k][j]) %= (p-1); return c; } Matrix power(Matrix a, ll b){ Matrix ret; ret.set(); while(b > 0){ if(b & 1)ret = ret * a; b >>= 1; a = a * a; }return ret; } ll power_mod(ll a, ll b){ ll ret = 1; while(b > 0){ if(b & 1)ret = ret * a % p; b >>= 1; a = a * a % p; }return ret; } int main(){ int test; scanf("%d", &test); while(test --){ cin >> n >> a >> b >> c >> p; if(n == 1){ cout << 1 % p << endl; continue; } if(a % p == 0){ cout << 0 << endl; continue; } mat.clear(); mat.a[0][0] = c, mat.a[1][0] = 1, mat.a[2][0] = b; mat.a[0][1] = 1; mat.a[2][2] = 1; ans.clear(); ans.a[0][0] = b, ans.a[0][1] = 0, ans.a[0][2] = 1; ans = ans * power(mat, n - 2); cout << power_mod(a, ans.a[0][0]) << endl; } return 0; }
建立两棵线段树跑分层图(据说要Dijkstra+Heap?)。注意第二棵的叶子节点向第一棵的叶子节点连边。
#include <algorithm> #include <iostream> #include <cstring> #include <cstdio> #include <vector> #include <queue> #define maxn 800010 using namespace std; int n, m, k; struct Edge{ int to, next, dis; }edge[maxn << 2]; int h[maxn], cnt, S, T, posT; void add(int u, int v, int d){ cnt ++; edge[cnt].to = v; edge[cnt].next = h[u]; edge[cnt].dis = d; h[u] = cnt; } vector<int> v1, v2; #define lc (id << 1) #define rc (id << 1 | 1) void build(int id, int l, int r){ if(l == r){ if(l == 1)S = id; posT = max(posT, id); return; } int mid = l + r >> 1; build(lc, l, mid); build(rc, mid+1, r); add(lc, id, 0); add(rc, id, 0); } void build2(int id, int l, int r){ if(l == r){ add(id + posT, id, 0); if(l == n)T = id + posT; return; } int mid = l + r >> 1; build2(lc, l, mid); build2(rc, mid+1, r); add(id + posT, lc + posT, 0); add(id + posT, rc + posT, 0); } void ask1(int id, int l, int r, int L, int R){ if(l == L && r == R){ v1.push_back(id); return; } int mid = l + r >> 1; if(R <= mid)ask1(lc, l, mid, L, R); else if(L > mid)ask1(rc, mid+1, r, L, R); else ask1(lc, l, mid, L, mid), ask1(rc, mid+1, r, mid+1, R); } void ask2(int id, int l, int r, int L, int R){ if(l == L && r == R){ v2.push_back(id); return; } int mid = l + r >> 1; if(R <= mid)ask2(lc, l, mid, L, R); else if(L > mid)ask2(rc, mid+1, r, L, R); else ask2(lc, l, mid, L, mid), ask2(rc, mid+1, r, mid+1, R); } queue<pair<int, int> > Q; int dis[maxn][11]; bool vis[maxn][11]; int main(){ int test; scanf("%d", &test); scanf("%d%d%d", &n, &m, &k); build(1, 1, n); build2(1, 1, n); int a, b, c, d, w, tot = posT << 1; for(int i = 1; i <= m; i ++){ scanf("%d%d%d%d%d", &a, &b, &c, &d, &w); v1.clear(), v2.clear(); ask1(1, 1, n, a, b); ask2(1, 1, n, c, d); ++ tot; for(int j = 0; j < v1.size(); j ++) add(v1[j], tot, 0); for(int j = 0; j < v2.size(); j ++) add(tot, v2[j] + posT, w); ++ tot; for(int j = 0; j < v2.size(); j ++) add(v2[j], tot, 0); for(int j = 0; j < v1.size(); j ++) add(tot, v1[j] + posT, w); } memset(dis, 0x7f, sizeof dis); Q.push(make_pair(S, 0)); dis[S][0] = 0; while(!Q.empty()){ int u = Q.front().first, k_ = Q.front().second; Q.pop(); vis[u][k_] = false; for(int i = h[u]; i; i = edge[i].next){ int v = edge[i].to; if(dis[v][k_] > dis[u][k_] + edge[i].dis){ dis[v][k_] = dis[u][k_] + edge[i].dis; if(!vis[v][k_])vis[v][k_] = true, Q.push(make_pair(v, k_)); } if(k_ < k && dis[v][k_+1] > dis[u][k_]){ dis[v][k_+1] = dis[u][k_]; if(!vis[v][k_+1])vis[v][k_+1] = true, Q.push(make_pair(v, k_+1)); } } } int ans = 0x7fffffff; for(int i = 0; i <= k; i ++) ans = min(ans, dis[T][i]); if(ans > 1e8)printf("CreationAugust is a sb!"); else printf("%d\n", ans); return 0; }
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原文地址:http://www.cnblogs.com/Candyouth/p/5400238.html