文章目录
- [1. 【2.3】P8784 [蓝桥杯 2022 省 B] 积木画](#1. 【2.3】P8784 [蓝桥杯 2022 省 B] 积木画)
- [2. 【2.4】P8656 [蓝桥杯 2017 国 B] 对局匹配](#2. 【2.4】P8656 [蓝桥杯 2017 国 B] 对局匹配)
- [3. 【2.5】[ABC365D] AtCoder Janken 3](#3. 【2.5】[ABC365D] AtCoder Janken 3)
- [4. 【2.6】P8703 [蓝桥杯 2019 国 B] 最优包含](#4. 【2.6】P8703 [蓝桥杯 2019 国 B] 最优包含)
- [5. 【2.7】P8624 [蓝桥杯 2015 省 AB] 垒骰子](#5. 【2.7】P8624 [蓝桥杯 2015 省 AB] 垒骰子)
- [6. 【2.8】P8774 [蓝桥杯 2022 省 A] 爬树的甲壳虫](#6. 【2.8】P8774 [蓝桥杯 2022 省 A] 爬树的甲壳虫)
- [7. 【2.9】[ARC189C] Balls and Boxes](#7. 【2.9】[ARC189C] Balls and Boxes)
正文
1. 【2.3】P8784 [蓝桥杯 2022 省 B] 积木画
【AC_Code】
cpp
#include <iostream>
#define IOS ios :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int mod = 1e9 + 7;
int N;
int main()
{
IOS; cin >> N;
if (N == 1)
{
cout << 1 << "\n";
return 0;
}
if (N == 2)
{
cout << 2 << "\n";
return 0;
}
if (N == 3)
{
cout << 5 << "\n";
return 0;
}
int a = 1, b = 2, c = 5, d;
for (int i = 4; i <= N; i ++)
{
d = c * 2 % mod + a;
a = b, b = c, c = d;
a %= mod, b %= mod, c %= mod, d %= mod;
}
cout << d << "\n";
return 0;
}2. 【2.4】P8656 [蓝桥杯 2017 国 B] 对局匹配
【AC_Code】
cpp
#include <iostream>
#define IOS ios :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int maxN = 1e5 + 10;
int N, K, index, ans, a[maxN];
int main()
{
IOS; cin >> N >> K;
for (int i = 0; i < N; i ++) cin >> index, a[index] ++;
if (K == 0)
{
for (int i = 0; i < N; i ++) if (a[i]) ans ++;
cout << ans << "\n";
return 0;
}
for (int i = 0; i < maxN; i ++)
{
if (a[i] < a[i + K]) a[i + K] -= a[i];
else a[i + K] = 0;
}
for (int i = 0; i < maxN - K; i ++) ans += a[i];
cout << ans << "\n";
return 0;
}3. 【2.5】[ABC365D] AtCoder Janken 3
【AC_Code】
cpp
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 2e5 + 10;
int n, ans, f[N][10];
string s;
int main()
{
cin >> n >> s;
if (s[0] == 'R') {
f[0][1] = 0;
f[0][2] = 1;
}
if (s[0] == 'P') {
f[0][2] = 0;
f[0][3] = 1;
}
if (s[0] == 'S') {
f[0][1] = 1;
f[0][3] = 0;
}
for (int i = 1; i < n; i++) {
if (s[i] == 'R') {
f[i][1] = max(f[i - 1][2], f[i - 1][3]);
f[i][2] = max(f[i - 1][1], f[i - 1][3]) + 1;
}
if (s[i] == 'P') {
f[i][2] = max(f[i - 1][1], f[i - 1][3]);
f[i][3] = max(f[i - 1][1], f[i - 1][2]) + 1;
}
if (s[i] == 'S') {
f[i][1] = max(f[i - 1][2], f[i - 1][3]) + 1;
f[i][3] = max(f[i - 1][1], f[i - 1][2]);
}
}
cout << max({f[n - 1][1], f[n - 1][2], f[n - 1][3]});
return 0;
}4. 【2.6】P8703 [蓝桥杯 2019 国 B] 最优包含
【AC_Code】
cpp
#include <iostream>
#include <cstring>
#include <algorithm>
#define IOS ios :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 1010;
int dp[maxn][maxn];
string S, T;
int main()
{
IOS; cin >> S >> T; S = " " + S, T = " " + T;
memset(dp, inf, sizeof(dp));
for (int i = 0; i < S.size() + 1; i ++) dp[i][0] = 0;
for (int i = 1; i <= S.size(); i ++) for (int j = 1; j <= S.size(); j ++)
{
if (S[i] == T[j]) dp[i][j] = dp[i-1][j-1];
else dp[i][j] = min(dp[i-1][j-1]+1, dp[i-1][j]);
}
cout << dp[S.size()][T.size()] << "\n";
return 0;
}5. 【2.7】P8624 [蓝桥杯 2015 省 AB] 垒骰子
【AC_Code】
cpp
#include<bits/stdc++.h>
using namespace std;
#define rep(x,y,z) for(int x=y;x<=z;x++)
typedef long long LL;
const int mod=1e9+7;
int n,m,a,b,oppo[7]={0,4,5,6,1,2,3};
bool st[7][7];
struct matrix{
LL c[7][7];
matrix(){memset(c,0,sizeof c);}
}A,res;
matrix operator * (matrix &x,matrix &y){
matrix t;
rep(i,1,6){
rep(j,1,6){
rep(k,1,6){
t.c[i][j]=(t.c[i][j]+x.c[i][k]*y.c[k][j])%mod;
}
}
}
return t;
}
void fastpow(LL k){
rep(i,1,6) res.c[1][i]=4;
rep(i,1,6){
rep(j,1,6){
if(st[i][oppo[j]]) A.c[i][j]=0;
else A.c[i][j]=4;
}
}
while(k){
if(k&1) res=res*A;
A=A*A;
k>>=1;
}
}
int main()
{
cin>>n>>m;
while(m--){
cin>>a>>b;
st[a][b]=st[b][a]=1;
}
fastpow(n-1);
LL ans=0;
rep(i,1,6) ans=(ans%mod+res.c[1][i]%mod)%mod;
cout<<ans;
return 0;
}6. 【2.8】P8774 [蓝桥杯 2022 省 A] 爬树的甲壳虫
【AC_Code】
cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int P = 998244353;
int n, a[N], b[N];
int fp(int x, int y) {
int res = 1;
for (; y; y >>= 1) {
if (y & 1) {
res = (1ll * res * x) % P;
}
x = (1ll * x * x) % P;
}
return res;
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d%d", &a[i], &b[i]);
}
int s1 = 1, s2 = 0, s3 = 0;
for (int i = 1; i <= n; i++) {
int p1 = (1ll * a[i] * fp(b[i], P - 2)) % P;
int p2 = (1ll * (b[i] - a[i]) * fp(b[i], P - 2)) % P;
s3 = (s3 + s1) % P;
s2 = (s2 + 1ll * s1 * p1) % P;
s1 = (1ll * s1 * p2) % P;
}
printf("%d", (1ll * s3 * fp(1 - s2 + P, P - 2)) % P);
return 0;
}7. 【2.9】[ARC189C] Balls and Boxes
【AC_Code】
cpp
#include <bits/stdc++.h>
#define F(i, a, b) for (int i = (a); i <= (b); ++i)
#define dF(i, a, b) for (int i = (a); i >= (b); --i)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<int, int> pii;
const int N = 2e5 + 5;
int n, x, a[N], b[N], p[N], q[N];
int m, k, s[N], t[N], tt[N];
bool vis[N];
int dp[N], c[N], ans;
void Update(int x, int k) {
while (x <= n) {
c[x] = max(c[x], k);
x += x & -x;
}
}
void Query(int &y, int x) {
while (x) {
y = max(y, c[x]);
x -= x & -x;
}
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
cin >> n >> x;
F(i, 1, n) {
cin >> a[i];
}
F(i, 1, n) {
cin >> b[i];
}
F(i, 1, n) {
cin >> p[i];
}
F(i, 1, n) {
cin >> q[i];
}
int u = p[x];
bool now = 0;
vis[x] = 1;
while (!vis[u]) {
now |= a[u];
vis[u] = 1;
if (now) {
s[++m] = u;
}
u = p[u];
}
F(i, 1, n) {
if (!vis[i] && a[i]) {
return cout << "-1", 0;
}
}
F(i, 1, n) {
vis[i] = 0;
}
u = q[x];
now = 0;
vis[x] = 1;
while (!vis[u]) {
now |= b[u];
vis[u] = 1;
if (now) {
t[++k] = u;
}
u = q[u];
}
F(i, 1, n) {
if (!vis[i] && b[i]) {
return cout << "-1", 0;
}
}
F(i, 1, k) {
tt[t[i]] = i;
}
F(i, 1, m) {
if (tt[s[i]]) {
Query(dp[i], tt[s[i]]);
Update(tt[s[i]], ++dp[i]);
}
}
F(i, 1, n) {
ans = max(ans, dp[i]);
}
cout << m + k - ans;
return 0;
}结语
感谢您的阅读！期待您的一键三连！欢迎指正！
