2020牛客暑期多校训练营（第九场）
A. Groundhog and 2-Power Representation
Groundhog and 2-Power Representation
⭐题意
将一个数表示成全部都时2的次方的形式，求这个数的值
⭐思路
一开始看范围就知道要么高精度，要么上py
而转念一想，不就是在
(前加一个**然后用py自带的
eval(evaluate)计算出来嘛🐎赛后看见一个这么短的代码
print(eval(input().replace('(','**(')))
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⭐代码
- py
s = input()
x = ''
for i in s:
if i == '(':
x += '**'
x += i
print(eval(x))
- C++
I. The Crime-solving Plan of Groundhog
The Crime-solving Plan of Groundhog
⭐题意
给一堆[0, 9]的数，拼接成两个数，使得乘积最小
⭐思路
先取这些数里面最小的一个非零数组成一个数，
再取剩下的组成一个最小的数
规则是，先在剩下的数取一个最小的一个非零数为第一位，
再在后面填0，然后从小到大开始往后填数字
这样可以保证乘积最小
⭐代码
- C++
/*************************************************************************
> FileName:
> Author: Lance
> Mail: lancelot_hcs@qq.com
> Date: 9102.1.8
> Description:
************************************************************************/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;
#define debug(a) cout << "*" << a << "*" << endl
const int INF = 0x3f3f3f3f;//int2147483647//ll9e18//unsigned ll 1e19
const int maxn = 100005;
//sacnf("%lf") printf("%f")
ll read() {
ll x = 0,f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
int t, n;
int a[100005];
int num[11];
vector<int> mul(vector<int> &A, int B) {
vector<int> C;
int t = 0;
for (int i = 0; i < A.size() || t; i++) {
if (i < A.size()) t += A[i] * B;
C.push_back(t % 10);
t /= 10;
}
reverse(C.begin(), C.end());
return C;
}
vector<int> A;
vector<int> ANS;
void solve() {
t = read();
while (t--) {
memset(num, 0, sizeof num);
A.clear();
ANS.clear();
n = read();
int min1 = INF;
for (int i = 1; i <= n; i++) {
a[i] = read();
if (a[i] != 0) min1 = min(min1, a[i]);
num[a[i]]++;
}
int b = min1;
num[b]--;
int fi = INF;
for (int i = 1; i <= 9; i++)
if (num[i] >= 1) {
fi = i;
break;
}
num[fi]--;
A.push_back(fi);
int con0 = num[0];
if (con0 > 0) while (con0--) A.push_back(0);
for (int i = 1; i <= 9; i ++) {
int con = num[i];
if (num[i] <= 0) continue;
while (con--) A.push_back(i);
}
reverse(A.begin(), A.end());
ANS = mul(A, b);
for (auto i : ANS) cout << i;
puts("");
}
}
int main()
{
// freopen("F:/Overflow/in.txt","r",stdin);
// ios::sync_with_stdio(false);
solve();
return 0;
}
- py
t=int(input())
for i in range(t):
n=int(input())
a=input().split()
a.sort()
pos=0
while a[pos]=='0':
pos=pos+1
x=a[pos]
y=a[pos+1]+'0'*pos+''.join(a[pos+2:])
print(int(x)*int(y))
c++选手骂骂咧咧退出群聊
join的用法，将一个容器，一般为列表或元组，用
''.join()引号的字符连接起来
如
>>> seq1 =['hello','good','boy','doiido']
>>> print' '.join(seq1)
hello good boy doiido
F. Groundhog Looking Dowdy
⭐题意
emmm
⭐思路
尺取，数据太水，
贪心都能过。。
⭐代码
/*************************************************************************
> FileName:
> Author: Lance
> Mail: lancelot_hcs@qq.com
> Date: 9102.1.8
> Description:
************************************************************************/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;
#define debug(a) cout << "*" << a << "*" << endl
const int INF = 0x3f3f3f3f;//int2147483647//ll9e18//unsigned ll 1e19
const int maxn = 4000005;
//sacnf("%lf") printf("%f")
ll read()
{
ll x = 0,f = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
int t, n, m, ans = INF, temp = 0, vis[maxn], all = 0;
struct node {
int day, dow;
bool operator < (const node& W) const {return dow < W.dow;}
}N[maxn];
void solve() {
n = read(), m = read();
for (int i = 1; i <= n; i++) {
int te = read();
for (int j = 1; j <= te; j++)
N[all].day = i, N[all++].dow = read();
}
sort(N, N + all);
for (int i = 0, j = 0; i < all; i++) {
if (vis[N[i].day] == 0) temp ++;
vis[N[i].day] ++;
while (temp >= m && j <= i) {
if ((--vis[N[j].day] == 0)) temp--;
ans = min(N[i].dow - N[j].dow, ans);
j++;
}
}
printf("%d", ans);
}
int main() {
// freopen("F:/Overflow/in.txt","r",stdin);
// ios::sync_with_stdio(false);
solve();
return 0;
}
E. Groundhog Chasing Death
⭐题意
给定x，y的值以及i，j的范围，求
\[\prod_{i=a}^{b} \prod_{j=c}^{d} g c d\left(x^{i}, y^{j}\right) \% \bmod \]
⭐思路
寻找规律发现，
gcd(a, b)实际上是a和b的公共质因数的乘积，而\(b^i\)为\(b\)的质因数个数的\(i\)倍，于是问题变成
⭐代码
/*************************************************************************
> FileName:
> Author: Lance
> Mail: lancelot_hcs@qq.com
> Date: 9102.1.8
> Description:
************************************************************************/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef __int128 LL;
typedef pair<int, int> PII;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 998244353;
#define debug(a) cout << "*" << a << "*" << endl
const int INF = 0x3f3f3f3f;//int2147483647//ll9e18//unsigned ll 1e19
const int maxn = 5000005;
//sacnf("%lf") printf("%f")
ll read()
{
ll x = 0,f = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
ll t, n;
ll top;
ll p[maxn], ans = 1;
ll num_x[maxn], num_y[maxn];
ll qpow(ll a, ll b) {
ll res = 1;
while (b) {
if (b & 1) res = (res * a) % mod;
a = (a * a) %mod;
b >>= 1;
}
return res;
}
void solve()
{
ll a = read(),b = read(),c = read(),d = read(),x = read(),y = read();
ll xx = x, yy = y, e = d - c + 1;
for (int i = 2; i <= xx / i; i++) {
if (xx % i == 0 && yy % i == 0) {
p[top] = i;
num_x[top] = num_y[top] = 0;
while (xx % i == 0) {
num_x[top]++;
xx /= i;
}
while (yy % i == 0) {
num_y[top]++;
yy /= i;
}
top++;
}
while (xx % i == 0) xx /= i;
while (yy % i == 0) yy /= i;
}
if (xx != 1 && yy != 1 && max(xx, yy) % min(xx, yy) == 0) {
ll i = min(xx, yy);
p[top] = i;
while (xx % i == 0) {
num_x[top]++;
xx /= i;
}
while (yy % i == 0) {
num_y[top]++;
yy /= i;
}
top++;
}
ans = 1;
for (ll i = 0; i < top; i++) {
ll z = 0;
ll temx, temy;
// cout << p[i] << ' ' << num_x[i] << ' ' << num_y[i] << endl;
for (ll j = a; j <= b; j++) {
temx = (num_x[i] * j) % mod;
temy = (num_y[i] * c) % mod;
if (temx <= temy) z += temx * e;
else if (temx >= (num_y[i] * d)) z += (num_y[i] * c + num_y[i] * d) * e / 2;
else {
ll r = temx / num_y[i];
z += (d - r) * temx;
z += (temy + r * num_y[i]) * (r - c + 1) / 2;
}
z %= mod - 1;
}
ans = (ans * qpow(p[i], z)) % mod;
}
printf("%lld\n", ans);
}
int main()
{
// freopen("F:/Overflow/in.txt","r",stdin);
// ios::sync_with_stdio(false);
solve();
return 0;
}
K.The Flee Plan of Groundhog
⭐题意
⭐思路
⭐代码
