2020省赛网赛

比赛链接

https://ac.nowcoder.com/acm/contest/15167

A. A Warm Welcome

题意

输出Shenzhen Institute of Computing Sciences

B. Mr.Maxwell and attractions

题意

你可以上午工作下午玩,也可以上午玩下午工作。

玩可以获得快乐,玩的时候有两类地方,一类是室内,一类是室外,室外下午玩会降低快乐值为\(80\%\),重复玩一个地方会导致快乐值降低\(60\%\), 可叠加。

你需要至少k个早上都在工作,问你最多获得多少快乐值。

题解

枚举玩多少次室内即可。用前缀和加速。

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#include<bits/stdc++.h>
using namespace std;
#pragma warning(disable:4996)
priority_queue<double, vector<double>, less<double>>ap, bp;
const int maxn = 1e5 + 100;
double a[maxn],b[maxn],at[maxn],bt[maxn],apreSum[maxn], bpreSum[maxn];
int n, m, t, k;
void initQ() {
int cnt = 1;
while (cnt <= t) {
double cura = ap.top();
ap.pop();
ap.push(cura * 0.6);
at[cnt] = cura;
apreSum[cnt] = apreSum[cnt - 1] + at[cnt];


double curb = bp.top();
bp.pop();
bp.push(curb * 0.6);
bt[cnt] = curb;
bpreSum[cnt] = bpreSum[cnt - 1] + bt[cnt];
cnt++;
}

}
void show() {
for (int i = 1; i <= t; i++) {
cout << at[i] << " " << bt[i] << endl;
}
for (int i = 1; i <= t; i++) {
cout << apreSum[i] << " " << bpreSum[i] << endl;
}
}
void solve() {
double ans = 0.0;
for (int x = 0; x <= t; x++) {
int y2 = t - x;
int y1 = y2;
if (x < k) {
y1 -= (k-x);
}
double curAns = apreSum[x] + bpreSum[y1] + (0.8 * (bpreSum[y2] - bpreSum[y1]));
ans = max(ans, curAns);
}
printf("%.2lf", ans);
}
int main() {
scanf("%d%d%d%d", &n, &m, &t, &k);
for (int i = 1; i <= n; i++) {
scanf("%lf", &a[i]);
ap.push(a[i]);
}

for (int i = 1; i <= m; i++) {
scanf("%lf", &b[i]);
bp.push(b[i]);
}
initQ();
solve();
}

C. Hamster and Equation

题意

输入n和k

输出 \[ x_1(x_1+1)+x_2(x_2+1)=k(x_3(x_3+1)+x_4(x_4+1)) \\ x_1,x_2,x_3,x_4 \in [-n,n] \] 的解的个数

数据范围

\(0\lt n,|k|\lt500\)

题解

预处理等式左边,枚举等式右边,复杂度\(n^2\)

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#include<bits/stdc++.h>
using namespace std;

#define ll long long
int main() {
int t; scanf("%d", &t);
while(t--) {
int n, k; scanf("%d%d", &n, &k);
unordered_map<ll, ll>ex;
for(int i = -n ; i <= n; ++i) {
for(int j = -n; j <= n; ++j) {
ll tem = 1ll * k * (1ll * i * (i + 1) + 1ll * j * (j + 1));
ex[tem]++;
}
}
ll cnt = 0;
for(int i = -n; i <= n; ++i) {
for(int j = -n; j <= n; ++j) {
ll tem = 1ll * i * (i + 1) + 1ll * j * (j + 1);
cnt += ex[tem];
}
}
printf("%lld\n", cnt);
}
}

D. WA

题意

输入一个字符串\(S\), 一个整数\(k\), 你可以修改字符串\(S\)的任意k个字母,问你修改后最多出现多少个\(aa\)子串。输出修改后的串。

数据范围

\(|S|\le 5\times10^5\)

\(k\le |S|\)

题解

预处理所有a之间的空隙,优先修改短的空隙,按顺序模拟即可。注意最后修改两端的空隙。

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#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int maxn = 5e5 + 5;
char s[maxn];

typedef pair<int, int> pii;

int main() {
int t;
//scanf("%d", &t);
t = 1;
while (t--) {
int n, k;
scanf("%d %d %s", &n, &k, s);

vector<pii> vec;
vec.push_back({int(1e9), -1});

for (int i = 0; i < n; i++) {
if (s[i] == 'a') {
vec.push_back({0, i});
} else {
vec.back().first++;
}
}

vector<pii> vec2;
for (auto x:vec) {
if (x.first != 0) {
vec2.push_back(x);
}
}


sort(vec2.begin(), vec2.end());

int cnt = 0;
for (auto p:vec2) {
if (p.second == -1) {
int up = p.second + 1;
while (up < n && s[up] != 'a') up++;
for (int i = up - 1; i > p.second; i--) {
if (cnt == k) {
break;
}
s[i] = 'a';
cnt++;

}
} else {
for (int i = p.second + 1; i < n && s[i] != 'a'; i++) {
if (cnt == k) {
break;
}
s[i] = 'a';
cnt++;

}
}

}

int base = 0;
for (int i = 0; i < n; i++) {
if (i != 0 && s[i] == 'a' && s[i - 1] == 'a') {
base++;
}
}


printf("%d\n%s\n", base, s);
}
}

E. Pipeline Maintenance

题意

给你一条长度为n的链,外加三个点,这三个点与链上每个点都连边,你得到了一个图,问你这个图的最小生成树的个数是多少。

输入只有一个n

数据范围

\(n\lt 10^9\)

题解

首先推出基尔霍夫矩阵,发现这个矩阵是有少量的地方有值,很明显他的行列式就是一个多项式,所以答案一定是一个多项式。

暴力计算前100项,然后BM线性递推即可。

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#include<bits/stdc++.h>
#pragma warning(disable:4996)
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1000000007;
ll powmod(ll a, ll b) { ll res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res * a % mod; a = a * a % mod; }return res; }
ll n;
namespace linear_seq {
const int N = 10010;
ll res[N], base[N], _c[N], _md[N];

vector<int> Md;
void mul(ll* a, ll* b, int k) {
rep(i, 0, k + k) _c[i] = 0;
rep(i, 0, k) if (a[i]) rep(j, 0, k) _c[i + j] = (_c[i + j] + a[i] * b[j]) % mod;
for (int i = k + k - 1; i >= k; i--) if (_c[i])
rep(j, 0, SZ(Md)) _c[i - k + Md[j]] = (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
rep(i, 0, k) a[i] = _c[i];
}
int solve(ll n, VI a, VI b) {
ll ans = 0, pnt = 0;
int k = SZ(a);
assert(SZ(a) == SZ(b));
rep(i, 0, k) _md[k - 1 - i] = -a[i]; _md[k] = 1;
Md.clear();
rep(i, 0, k) if (_md[i] != 0) Md.push_back(i);
rep(i, 0, k) res[i] = base[i] = 0;
res[0] = 1;
while ((1ll << pnt) <= n) pnt++;
for (int p = pnt; p >= 0; p--) {
mul(res, res, k);
if ((n >> p) & 1) {
for (int i = k - 1; i >= 0; i--) res[i + 1] = res[i]; res[0] = 0;
rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
}
}
rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
if (ans < 0) ans += mod;
return ans;
}
VI BM(VI s) {
VI C(1, 1), B(1, 1);
int L = 0, m = 1, b = 1;
rep(n, 0, SZ(s)) {
ll d = 0;
rep(i, 0, L + 1) d = (d + (ll)C[i] * s[n - i]) % mod;
if (d == 0) ++m;
else if (2 * L <= n) {
VI T = C;
ll c = mod - d * powmod(b, mod - 2) % mod;
while (SZ(C) < SZ(B) + m) C.pb(0);
rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
L = n + 1 - L; B = T; b = d; m = 1;
}
else {
ll c = mod - d * powmod(b, mod - 2) % mod;
while (SZ(C) < SZ(B) + m) C.pb(0);
rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
++m;
}
}
return C;
}
int gao(VI a, ll n) {
VI c = BM(a);
c.erase(c.begin());
rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
}
};
int a[1000] = { 1,20,216,1840,13775,95040,619801,3878720,23520456,139127500,806585879,599175652,861664394,707058859,417979870,901047604,478633297,859865743,368755586,930893321,243990638,416220770,156922876,768961406,372030171,188255286,753829864,246844887,442658427,357182332,744405222,783203806,469197530,863684841,605924134,166060944,506226150,446220745,171110722,498919220,700717610,739340306,607058637,253306001,703467596,231535400,903802311,143421365,864786702,113238066,748503739,575557576,596128329,62322981,98752077,240806338,956345596,374036254,976624372,344168146,879827644,658625868,76392155,576562868,336205776,392396240,70109394,71982377,780620194,821250696,668859101,16081127,485315931,278337560,180126339,172842175,402815218,33449281,512582468,457919375,64916357,966658493,531395887,571188277,243742869,586283678,302575818,40249574,901283990,633872644,396221397,13159314,543397157,575791218,993120783,494677489,620570286,883513941,153287837,309800837 };
int main() {
vector<int>v;
for (int i = 0; i < 50; i++) {
v.push_back(a[i]);
}
scanf("%lld", &n);
printf("%lld\n", 1LL * linear_seq::gao(v, n - 1) % mod);
}

F. Meet in another world, enjoy tasty food!

题意

\(n\)个人在排队,给你长度为n的数组,这是每个人都的耐心值,排名为\(i\)的人每秒会丧失\(i\)点耐心,当耐心值低于\(0\)的时候,这个人会离开队列,与此同时,他后面的人的排名都会减少1。你需要输出出队顺序。

数据范围

\(n\lt 1000\)

\(a_i<10^{18}\)

题解

暴力计算每一轮谁离开了队列。

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#include<bits/stdc++.h>

using namespace std;

#define ll long long
typedef pair<ll, ll> pll;

int main() {
ll t;
t = 1;
while (t--) {
ll n;
scanf("%lld", &n);
vector<pll> a;

for (ll i = 1; i <= n; i++) {
ll x;
scanf("%lld", &x);
a.push_back(make_pair(x, i));
}

vector<ll> ans;
for (ll _ = 1; _ <= n; _++) {
ll k = 1e18 + 100;
for (ll i = 0; i < a.size(); i++) {
ll rank = i + 1;
k = min(k, (a[i].first + rank - 1) / rank);
}
vector<pll> b;

if (k != 1) {
k--;
for (ll i = 0, ii = 1; i < a.size(); i++) {
if (a[i].first - ii * k <= 0) {
ans.push_back(a[i].second);
} else {
a[i].first -= k * ii;
b.push_back(a[i]);
ii++;
}
}
a = b;
b.clear();
}

k = 1;
for (ll i = 0, ii = 1; i < a.size(); i++) {
if (a[i].first - ii * k <= 0) {
ans.push_back(a[i].second);
} else {
a[i].first -= k * ii;
b.push_back(a[i]);
ii++;
}
}
a = b;
b.clear();
}

for (ll x: ans) {
printf("%lld ", x);
}
printf("\n");
}
}