題目大意
將範圍為 $1 \sim 100$ 的數字排序並輸出。時限很緊,要優化 IO。
題解
排序小範圍的數字可以用 Counting Sort,請參考【題解】Zerojudge c431 - Sort ! Sort ! Sort ! 的題解。IO 優化的相關技巧請參考我的【筆記】IO 優化這篇文章,以下的程式碼是使用該文章中所提到的 tourist IO。
#include <bits/stdc++.h>
using namespace std;
namespace std {
static struct FastInput {
static constexpr int BUF_SIZE = 1 << 20;
char buf[BUF_SIZE];
size_t chars_read = 0;
size_t buf_pos = 0;
FILE *in = stdin;
char cur = 0;
inline char get_char() {
if(buf_pos >= chars_read) {
chars_read = fread(buf, 1, BUF_SIZE, in);
buf_pos = 0;
buf[0] = (chars_read == 0 ? -1 : buf[0]);
}
return cur = buf[buf_pos++];
// return cur = getchar_unlocked();
}
inline void tie(int) {}
inline explicit operator bool() { return cur != -1; }
inline static bool is_blank(char c) { return c <= ' '; }
inline bool skip_blanks() {
while(is_blank(cur) && cur != -1) {
get_char();
}
return cur != -1;
}
inline FastInput& operator>>(char& c) {
skip_blanks();
c = cur;
return *this;
}
inline FastInput& operator>>(string& s) {
if(skip_blanks()) {
s.clear();
do {
s += cur;
} while(!is_blank(get_char()));
}
return *this;
}
template<class T>
inline FastInput& read_integer(T& n) {
n = 0;
if(skip_blanks()) {
int sign = +1;
if(cur == '-') {
sign = -1;
get_char();
}
do {
n += n + (n << 3) + cur - '0';
} while(!is_blank(get_char()));
n *= sign;
}
return *this;
}
template<class T> inline typename enable_if<is_integral<T>::value, FastInput&>::type operator>>(T& n) { return read_integer(n); }
inline FastInput& operator>>(__int128& n) { return read_integer(n); }
template<class T>
inline typename enable_if<is_floating_point<T>::value, FastInput&>::type operator>>(T& n) {
n = 0;
if(skip_blanks()) {
string s;
(*this) >> s;
sscanf(s.c_str(), "%lf", &n);
}
return *this;
}
} fast_input;
#define istream FastInput
#define cin fast_input
static struct FastOutput {
static constexpr int BUF_SIZE = 1 << 20;
char buf[BUF_SIZE];
size_t buf_pos = 0;
static constexpr int TMP_SIZE = 1 << 20;
char tmp[TMP_SIZE];
FILE *out = stdout;
inline void put_char(char c) {
// buf[buf_pos++] = c;
// if(buf_pos == BUF_SIZE) {
// fwrite(buf, 1, buf_pos, out);
// buf_pos = 0;
// }
putchar_unlocked(c);
}
~FastOutput() {
// fwrite(buf, 1, buf_pos, out);
}
inline FastOutput& operator<<(char c) {
put_char(c);
return *this;
}
inline FastOutput& operator<<(const char* s) {
while(*s) {
put_char(*s++);
}
return *this;
}
inline FastOutput& operator<<(const string& s) {
for(int i = 0; i < (int) s.size(); i++) {
put_char(s[i]);
}
return *this;
}
template<class T>
inline char* integer_to_string(T n) {
char* p = tmp + TMP_SIZE - 1;
if(n == 0) {
*--p = '0';
} else {
bool is_negative = false;
if(n < 0) {
is_negative = true;
n = -n;
}
while(n > 0) {
*--p = (char) ('0' + n % 10);
n /= 10;
}
if(is_negative) {
*--p = '-';
}
}
return p;
}
template<class T> inline typename enable_if<is_integral<T>::value, char*>::type stringify(T n) { return integer_to_string(n); }
inline char* stringify(__int128 n) { return integer_to_string(n); }
template<class T>
inline typename enable_if<is_floating_point<T>::value, char*>::type stringify(T n) {
sprintf(tmp, "%.17f", n);
return tmp;
}
template<class T>
inline FastOutput& operator<<(const T& n) {
auto p = stringify(n);
for(; *p != 0; p++) {
put_char(*p);
}
return *this;
}
} fast_output;
#define ostream FastOutput
#define cout fast_output
} // namespace std
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
vector<int> cnt(101);
int x;
while(cin >> x) {
cnt[x]++;
}
for(int i = 1; i <= 100; ++i) {
while(cnt[i]--) {
cout << i << " ";
}
}
cout << "\n";
return 0;
}
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© 若無特別註明,本站文章皆由 WeakMouse's Coding Blog 原創 ,轉載引用本文前請先留言告知。本文轉載請註明文章源自 WeakMouse's Coding Blog ,作者 ,並附上原文連結: 【題解】Zerojudge f282 - 松鼠的願望
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