给出的仅为我的做题顺序，与题目难度不一定匹配
F、排列计算
前缀和
给定区间，可以控制区间元素加1，最后把区间元素的染色次数排序，从次数最大开始填最大的数。
具体区间元素+1，又不用模拟去实现，直接在左端点处加1，右端点后面一个点减1即可。最后求前缀和就行。
再把求好的前缀和数组排个序，大功告成
#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize(2)
#pragma GCC optimize(3)
typedef long long ll;
#define INF 0x3f3f3f3f
const int mod = 1e9+7;
const int maxn = 8000*8000+5;
#define iss ios::sync_with_stdio(false)
inline ll read(){
ll s = 0, w = 1; char ch = getchar();
while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }
while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
return s * w;
}
ll num[200005];
int main(){
int n = read(),m = read();
while(m--){
int l = read(),r = read();
num[l]++;
num[r+1]--;
}
for(int i = 0 ; i <= n ; ++i ) num[i]+=num[i-1];
sort(num+1,num+1+n);
ll ans = 0;
for(int i = 1 ; i <= n ;++i){
ans += (i*num[i]);
}
cout<<ans<<endl;
} B、伤害计算
写法，用了个
和异常处理。对输入的字符串进行一个个数据处理即可，注意最后判断整数精度别开太小
）我竟然被1e-6坑了一发……
当然还有更好的写法，直接乘个2去算，最后判奇偶就行了。
import math
a = list(input().split('+'))
ans = 0.0
for i in a:
try:
tmp = int(i)
ans += tmp
except:
cnt,tmp = map(int,i.split('d'))
ans += cnt*(tmp+1)*0.5
if ans-int(ans)>0.1:
print(ans)
else :
ans=int(ans)
print(ans)
A、张老师和菜哭武的游戏
前导知识：拓展欧里几得
我们知道对于 存在整数解x与y，需要gcd(a,b)是c的因子。
对于用a和b求解出来的数随意组合也符合上面的
那么这题，给出了a，b，直接求解gcd即可，用n除掉gcd就是可以拿走的数。在对这个数判断奇偶即可
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
inline int read() {
int s = 0, w = 1; char ch = getchar();
while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }
while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
return s * w;
}
int main() {
int T = read();
while (T--) {
int n = read(), a = read(), b = read();
int tmp = __gcd(a, b);
if (tmp != 1) n /= tmp;
if (n & 1) puts("Yes");
else puts("No");
}
return 0;
}
D、车辆调度
地图很小，车辆很少，直接暴力破解即可。
直接跑地图即可，再把每个位置上下左右都走一遍
(钟涛大佬tql)
#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize(2)
#pragma GCC optimize(3)
typedef long long ll;
#define INF 0x3f3f3f3f
const int mod = 1e9 + 7;
const int maxn = 8000 * 8000 + 5;
#define iss ios::sync_with_stdio(false)
inline ll read() {
ll s = 0, w = 1; char ch = getchar();
while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }
while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
return s * w;
}
int w, h, kk;
char Map[15][15];
bool dfs(int x) {
if (x > kk) return false;
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
if (Map[i][j] == 'R') {
Map[i][j] = '.';
int k;
if (i > 0 && Map[i - 1][j] == '.' || Map[i - 1][j] == 'D') {
for (int k = i - 1; k >= 0; k--)
if (k == 0 || Map[k - 1][j] == 'R' || Map[k - 1][j] == 'X') {
if (Map[k][j] == 'D') return true;
Map[k][j] = 'R';
if (dfs(x + 1)) return true;
Map[k][j] = '.';
break;
}
}
if (i + 1 < h && Map[i + 1][j] == '.' || Map[i + 1][j] == 'D') {
for (int k = i + 1; k < h; ++k) {
if (k == h - 1 || Map[k + 1][j] == 'R' || Map[k + 1][j] == 'X') {
if (Map[k][j] == 'D') return true;
Map[k][j] = 'R';
if (dfs(x + 1)) return true;
Map[k][j] = '.';
break;
}
}
}
if (j + 1 < w && Map[i][j + 1] == '.' || Map[i][j + 1] == 'D') {
for (int k = j + 1; k < w; ++k) {
if (k == w - 1 || Map[i][k + 1] == 'R' || Map[i][k + 1] == 'X') {
if (Map[i][k] == 'D') return true;
Map[i][k] = 'R';
if (dfs(x + 1)) return true;
Map[i][k] = '.';
break;
}
}
}
if (j > 0 && Map[i][j - 1] == '.' || Map[i][j - 1] == 'D') {
for (int k = j - 1; k >= 0; k--)
if (k == 0 || Map[i][k - 1] == 'R' || Map[i][k - 1] == 'X') {
if (Map[i][k] == 'D') return true;
Map[i][k] = 'R';
if (dfs(x + 1)) return true;
Map[i][k] = '.';
break;
}
}
Map[i][j] = 'R';
}
}
}
return false;
}
int main() {
w = read(), h = read(), kk = read();
for (int i = 0; i < h; ++i)
scanf("%s", Map[i]);
if (dfs(1)) puts("YES");
else puts("NO");
}
好了下面是比赛时间未写出来的了
E、弦
分子式卡特兰数我看出来了……居然死在了分母手上，我一直以为分母是个末项(n * 2 - 1)的等差数列……我一直只连接2条线。
现在发现是多蠢。分母应该是
再结合卡特兰数的通项公式
约分化简得到答案等于 在处理一下阶乘数组，利用费马小定理求解乘法逆元就是答案了。
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
inline int read() {
int s = 0, w = 1; char ch = getchar();
while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }
while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
return s * w;
}
const int MOD = 1e9 + 7;
const int N = 1e7 + 7;
ll jc[N];
ll qpow(ll a, ll b) {
ll ans = 1;
while (b) {
if (b & 1) (ans *= a) %= MOD;
b >>= 1;
(a *= a) %= MOD;
}
return ans;
}
int main() {
int n = read();
jc[0] = 1;
for (int i = 1; i <= n + 1; ++i) jc[i] = jc[i - 1] * i % MOD;
printf("%lld\n", qpow(2,n) * qpow(jc[n + 1], MOD - 2) % MOD);
return 0;
}
F、斐波那契和
我敢保证这题是最憋屈的，以为两天前的牛客练习赛63，斐波那契字符串刚刚好说了一句杜教BM，本菜鸡只百度了一下，知道是个帮你求递推式的没有去记，因为想着有OEIS，如果知道前几项直接求道通项公式就行了……然后结果就是这题没法做。。
具体实现……就是套板子，一般来说求解到前10项左右即可出答案，稳健一点，直接开了10000的数组，全部放进去。算到答案即可。
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
typedef long long ll;
#define SZ(x) ((ll)(x).size())
typedef vector<ll> VI;
typedef pair<ll, ll> PII;
const ll mod = 998244353;
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; }
// head
ll _, n;
namespace linear_seq {
const ll N = 10010;
ll res[N], base[N], _c[N], _md[N];
vector<ll> Md;
void mul(ll* a, ll* b, ll 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 (ll 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];
}
ll solve(ll n, VI a, VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
// printf("%d\n",SZ(b));
ll ans = 0, pnt = 0;
ll 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 (ll p = pnt; p >= 0; p--) {
mul(res, res, k);
if ((n >> p) & 1) {
for (ll 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);
ll 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;
}
ll 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)));
}
};
inline ll read() {
ll s = 0, w = 1; char ch = getchar();
while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }
while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
return s * w;
}
ll f[10005];
int main() {
ll n = read(), k = read();
f[1] = 1, f[2] = 1;
for (int i = 3; i <= 10000; ++i)
f[i] = (f[i - 1] + f[i - 2]) % mod;
VI a;
ll x = 0;
for (int i = 1; i <= 10000; ++i) {
x = (x + powmod(i, k) * f[i] % mod) % mod;
a.push_back(x);
}
printf("%lld\n", linear_seq::gao(a, n - 1));
}
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