杂七杂八的知识

1.重载运算符

理解并不难，主要是格式值得注意。
另外要注意在队列中要注意排序的顺序，q.top()/q.front() 访问的是队首元素，如果queue是升序，q.front()就是最大的；

include

include

using namespace std;

struct student {
string name;
int score;
};

struct cmp {
bool operator()(const student& a, const student& b) const {
return a.score < b.score || (a.score == b.score && a.name > b.name);
}
};

priority_queue<student, vector, cmp> pq;
int main() {
return 0;
}

2.sort

当用sort排序vector时，这样写：

sort(vector.begin() , vector.end() , cmp);

3.高精度计算器

模板好长啊qwq

include

include

static const int LEN = 1004;

int a[LEN], b[LEN], c[LEN], d[LEN];

void clear(int a[]) {
for (int i = 0; i < LEN; ++i) a[i] = 0;
}

void read(int a[]) {
static char s[LEN + 1];
scanf("%s", s);

clear(a);

int len = strlen(s);
for (int i = 0; i < len; ++i) a[len - i - 1] = s[i] - '0';
}

void print(int a[]) {
int i;
for (i = LEN - 1; i >= 1; --i)
if (a[i] != 0) break;
for (; i >= 0; --i) putchar(a[i] + '0');
putchar('\n');
}

void add(int a[], int b[], int c[]) {
clear(c);

for (int i = 0; i < LEN - 1; ++i) {
c[i] += a[i] + b[i];
if (c[i] >= 10) {
c[i + 1] += 1;
c[i] -= 10;
}
}
}

void sub(int a[], int b[], int c[]) {
clear(c);

for (int i = 0; i < LEN - 1; ++i) {
c[i] += a[i] - b[i];
if (c[i] < 0) {
c[i + 1] -= 1;
c[i] += 10;
}
}
}

void mul(int a[], int b[], int c[]) {
clear(c);

for (int i = 0; i < LEN - 1; ++i) {
for (int j = 0; j <= i; ++j) c[i] += a[j] * b[i - j];

if (c[i] >= 10) {
c[i + 1] += c[i] / 10;
c[i] %= 10;
}
}
}

bool greater_eq(int a[], int b[], int last_dg, int len) {
if (a[last_dg + len] != 0) return true;
for (int i = len - 1; i >= 0; --i) {
if (a[last_dg + i] > b[i]) return true;
if (a[last_dg + i] < b[i]) return false;
}
return true;
}

void div(int a[], int b[], int c[], int d[]) {
clear(c);
clear(d);

int la, lb;
for (la = LEN - 1; la > 0; --la)
if (a[la - 1] != 0) break;
for (lb = LEN - 1; lb > 0; --lb)
if (b[lb - 1] != 0) break;
if (lb == 0) {
puts("> <");
return;
}

for (int i = 0; i < la; ++i) d[i] = a[i];
for (int i = la - lb; i >= 0; --i) {
while (greater_eq(d, b, i, lb)) {
for (int j = 0; j < lb; ++j) {
d[i + j] -= b[j];
if (d[i + j] < 0) {
d[i + j + 1] -= 1;
d[i + j] += 10;
}
}
c[i] += 1;
}
}
}

int main() {
read(a);

char op[4];
scanf("%s", op);

read(b);

switch (op[0]) {
case '+':
add(a, b, c);
print(c);
break;
case '-':
sub(a, b, c);
print(c);
break;
case '*':
mul(a, b, c);
print(c);
break;
case '/':
div(a, b, c, d);
print(c);
print(d);
break;
default:
puts("> <");
}

return 0;
}

最长上升子序列 $ O(nlogn) $ 的做法

挺难想的 ， 我觉得考场现写俺也调不出来

include <bits/stdc++.h>

using namespace std;

const int N = 1e5+10;
int n,len;
int a[N],stk[N];

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n;
for(int i=1 ; i<=n ; i++) cin >> a[i];

stk[0] = -2e9 ;
for(int i=1 ; i<=n ; i++) {
if(a[i] > stk[len]) stk[++len] = a[i];
else stk[lower_bound(stk+1 , stk+len+1 , a[i]) - stk] = a[i];
}

cout << len << endl;
return 0;
}