Bestcoder Round# 80

[1003 Sequence]

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;
}

　　

[1005 Road]

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;
}