A. Groundhog and 2-Power Representation（eval应用）
题意：2(2)表示2^2 = 4，计算给出的表达式
eval函数计算表达式值，值得注意的是表达式中需要用**来表示乘方。一行过
搞了37分钟，python还是不熟啊
print(eval(str(input()).replace('(', '**(')))F. Groundhog Looking Dowdy（尺取）
题意：
共 n 天，每天有一些数量的衣服，每件衣服有个邋遢值，选取 m 天，m 天里每天选取一件衣服穿，最小化这 m 件衣服邋遢值的最大值和最小值之间的差值
思路：
把所有衣服按邋遢值排序，对应出每件衣服所代表的天数，假如区间 [l, r] ***有 m 个不同的天数，则差值为 s[r].a - s[l].a，尺取解决，快读快写可优化200ms
尺取也不熟啊
scanf + printf：
+ 快读写：
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int N = 1e6 + 5;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while(!isdigit(ch)) {
if(ch == '-')
f = -1;
ch = getchar();
}
while(isdigit(ch)) {
x = (x << 1) + (x << 3) + (ch ^ 48);
ch = getchar();
}
return x * f;
}
inline void write(int x)
{
if(x < 0) x = -x, putchar('-');
if(x > 9) write(x / 10);
putchar(x % 10 + '0');
}
struct node
{
int a, id;
} s[2 * N];
bool cmp(node x, node y) {
return x.a <= y.a;
}
int mp[N];
int main()
{
for(int i = 0; i < N; ++i) mp[i] = 0;
int n, m, k, tot = 0;
n = read(), m = read();
for(int i = 1; i <= n; ++i) {
k = read();
while(k--) {
s[++tot].a = read();
s[tot].id = i;
}
}
sort(s + 1, s + tot + 1, cmp);
int l = 1, r = 0, cnt = 0;
int minn = inf;
while(1) {
while(r < tot && cnt < m) {
r++;
if(!mp[s[r].id]) cnt++;
mp[s[r].id]++;
}
if(cnt < m) break;
minn = min(minn, s[r].a - s[l].a);
mp[s[l].id]--;
if(!mp[s[l].id]) cnt--;
l++;
}
write(minn), printf("\n");
return 0;
}
I. The Crime-solving Plan of Groundhog（思维 + 大数）
题意：
给出 n 个数字（0~9），用它们构造两个没有前导零的正整数，求这两个数的最小积
思路：跟着感觉走，最小积肯定是拿出一个最小的数字来和其他的乘了，题解给的证明：
套的模板，先处理出每个数字的个数再加，排序的话T了
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int N = 1e5 + 5;
//大数
struct BigInteger
{
static const int BASE = 100000000; //和WIDTH保持一致
static const int WIDTH = 8; //八位一存储,如修改记得修改输出中的%08d
bool sign; //符号, 0表示负数
size_t length; //位数
vector<int> num; //反序存
//构造函数
BigInteger(long long x = 0)
{
*this = x;
}
BigInteger(const string &x)
{
*this = x;
}
BigInteger(const BigInteger &x)
{
*this = x;
}
//剪掉前导0，并且求一下数的位数
void cutLeadingZero()
{
while (num.back() == 0 && num.size() != 1)
{
num.pop_back();
}
int tmp = num.back();
if (tmp == 0)
{
length = 1;
}
else
{
length = (num.size() - 1) * WIDTH;
while (tmp > 0)
{
length++;
tmp /= 10;
}
}
}
//赋值运算符
BigInteger &operator=(long long x)
{
num.clear();
if (x >= 0)
{
sign = true;
}
else
{
sign = false;
x = -x;
}
do
{
num.push_back(x % BASE);
x /= BASE;
}
while (x > 0);
cutLeadingZero();
return *this;
}
BigInteger &operator=(const string &str)
{
num.clear();
sign = (str[0] != '-'); //设置符号
int x, len = (str.size() - 1 - (!sign)) / WIDTH + 1;
for (int i = 0; i < len; i++)
{
int End = str.size() - i * WIDTH;
int start = max((int)(!sign), End - WIDTH); //防止越界
sscanf(str.substr(start, End - start).c_str(), "%d", &x);
num.push_back(x);
}
cutLeadingZero();
return *this;
}
BigInteger &operator=(const BigInteger &tmp)
{
num = tmp.num;
sign = tmp.sign;
length = tmp.length;
return *this;
}
//绝对值
BigInteger abs() const
{
BigInteger ans(*this);
ans.sign = true;
return ans;
}
//正号
const BigInteger &operator+() const
{
return *this;
}
//负号
BigInteger operator-() const
{
BigInteger ans(*this);
if (ans != 0)
ans.sign = !ans.sign;
return ans;
}
// + 运算符
BigInteger operator+(const BigInteger &b) const
{
if (!b.sign)
{
return *this - (-b);
}
if (!sign)
{
return b - (-*this);
}
BigInteger ans;
ans.num.clear();
for (int i = 0, g = 0;; i++)
{
if (g == 0 && i >= num.size() && i >= b.num.size())
break;
int x = g;
if (i < num.size())
x += num[i];
if (i < b.num.size())
x += b.num[i];
ans.num.push_back(x % BASE);
g = x / BASE;
}
ans.cutLeadingZero();
return ans;
}
// - 运算符
BigInteger operator-(const BigInteger &b) const
{
if (!b.sign)
{
return *this + (-b);
}
if (!sign)
{
return -((-*this) + b);
}
if (*this < b)
{
return -(b - *this);
}
BigInteger ans;
ans.num.clear();
for (int i = 0, g = 0;; i++)
{
if (g == 0 && i >= num.size() && i >= b.num.size())
break;
int x = g;
g = 0;
if (i < num.size())
x += num[i];
if (i < b.num.size())
x -= b.num[i];
if (x < 0)
{
x += BASE;
g = -1;
}
ans.num.push_back(x);
}
ans.cutLeadingZero();
return ans;
}
// * 运算符
BigInteger operator*(const BigInteger &b) const
{
int lena = num.size(), lenb = b.num.size();
BigInteger ans;
for (int i = 0; i < lena + lenb; i++)
ans.num.push_back(0);
for (int i = 0, g = 0; i < lena; i++)
{
g = 0;
for (int j = 0; j < lenb; j++)
{
long long x = ans.num[i + j];
x += (long long)num[i] * (long long)b.num[j];
ans.num[i + j] = x % BASE;
g = x / BASE;
ans.num[i + j + 1] += g;
}
}
ans.cutLeadingZero();
ans.sign = (ans.length == 1 && ans.num[0] == 0) || (sign == b.sign);
return ans;
}
//*10^n 大数除大数中用到
BigInteger e(size_t n) const
{
int tmp = n % WIDTH;
BigInteger ans;
ans.length = n + 1;
n /= WIDTH;
while (ans.num.size() <= n)
ans.num.push_back(0);
ans.num[n] = 1;
while (tmp--)
ans.num[n] *= 10;
return ans * (*this);
}
// /运算符 (大数除大数)
BigInteger operator/(const BigInteger &b) const
{
BigInteger aa((*this).abs());
BigInteger bb(b.abs());
if (aa < bb)
return 0;
char *str = new char[aa.length + 1];
memset(str, 0, sizeof(char) * (aa.length + 1));
BigInteger tmp;
int lena = aa.length, lenb = bb.length;
for (int i = 0; i <= lena - lenb; i++)
{
tmp = bb.e(lena - lenb - i);
while (aa >= tmp)
{
str[i]++;
aa = aa - tmp;
}
str[i] += '0';
}
BigInteger ans(str);
delete[] str;
ans.sign = (ans == 0 || sign == b.sign);
return ans;
}
// %运算符
BigInteger operator%(const BigInteger &b) const
{
return *this - *this / b * b;
}
// ++ 运算符
BigInteger &operator++()
{
*this = *this + 1;
return *this;
}
// -- 运算符
BigInteger &operator--()
{
*this = *this - 1;
return *this;
}
// += 运算符
BigInteger &operator+=(const BigInteger &b)
{
*this = *this + b;
return *this;
}
// -= 运算符
BigInteger &operator-=(const BigInteger &b)
{
*this = *this - b;
return *this;
}
// *=运算符
BigInteger &operator*=(const BigInteger &b)
{
*this = *this * b;
return *this;
}
// /= 运算符
BigInteger &operator/=(const BigInteger &b)
{
*this = *this / b;
return *this;
}
// %=运算符
BigInteger &operator%=(const BigInteger &b)
{
*this = *this % b;
return *this;
}
// < 运算符
bool operator<(const BigInteger &b) const
{
if (sign != b.sign) //正负，负正
{
return !sign;
}
else if (!sign && !b.sign) //负负
{
return -b < -*this;
}
//正正
if (num.size() != b.num.size())
return num.size() < b.num.size();
for (int i = num.size() - 1; i >= 0; i--)
if (num[i] != b.num[i])
return num[i] < b.num[i];
return false;
}
bool operator>(const BigInteger &b) const
{
return b < *this; // > 运算符
}
bool operator<=(const BigInteger &b) const
{
return !(b < *this); // <= 运算符
}
bool operator>=(const BigInteger &b) const
{
return !(*this < b); // >= 运算符
}
bool operator!=(const BigInteger &b) const
{
return b < *this || *this < b; // != 运算符
}
bool operator==(const BigInteger &b) const
{
return !(b < *this) && !(*this < b); //==运算符
}
// 逻辑运算符
bool operator||(const BigInteger &b) const
{
return *this != 0 || b != 0; // || 运算符
}
bool operator&&(const BigInteger &b) const
{
return *this != 0 && b != 0; // && 运算符
}
bool operator!()
{
return (bool)(*this == 0); // ！ 运算符
}
//重载<<使得可以直接输出大数
friend ostream &operator<<(ostream &out, const BigInteger &x)
{
if (!x.sign)
out << '-';
out << x.num.back();
for (int i = x.num.size() - 2; i >= 0; i--)
{
char buf[10];
//如WIDTH和BASR有变化,此处要修改为%0(WIDTH)d
sprintf(buf, "%08d", x.num[i]);
for (int j = 0; j < strlen(buf); j++)
out << buf[j];
}
return out;
}
//重载>>使得可以直接输入大数
friend istream &operator>>(istream &in, BigInteger &x)
{
string str;
in >> str;
size_t len = str.size();
int start = 0;
if (str[0] == '-')
start = 1;
if (str[start] == '\0')
return in;
for (int i = start; i < len; i++)
{
if (str[i] < '0' || str[i] > '9')
return in;
}
x.sign = !start;
x = str.c_str();
return in;
}
};
inline int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
return x*f;
}
int a[N], cnt[10];
int main()
{
int t;
t = read();
while(t--) {
int n;
n = read();
for(int i = 0; i < 10; ++i) cnt[i] = 0;
for(int i = 0;i < n;i ++) {
a[i] = read();
cnt[a[i]]++;
}
BigInteger mi = 0;
for(int i = 1; i < 10; ++i) {
if(cnt[i]) {
mi = i;
cnt[i]--;
break;
}
}
string s = "";
for(int i = 1; i < 10; ++i) {
if(cnt[i]) {
s += i + '0';
cnt[i]--;
break;
}
}
for(int i = 0; i < 10; ++i)
for(int j = 1; j <= cnt[i]; ++j)
s += i + '0';
BigInteger sum = s;
BigInteger ans = mi * sum;
cout<<ans<<'\n';
}
return 0;
}
明天再补zzzzz
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