第三次双周赛

T1

map的简单运用，或者读进来再一个一个判断就行。

#include<bits/stdc++.h>
using namespace std;
map<char ,int> m;
int a[9];
int main () {
   m['1'] = 1; 
   m['A'] = 1;
   m['Q'] = 1;
   m['Z'] = 1;
   m['2'] = 2;
   m['W'] = 2;
   m['S'] = 2;
   m['X'] = 2;
   m['3'] = 3;
   m['E'] = 3;
   m['D'] = 3;
   m['C'] = 3;
   m['R'] = 4;
   m['4'] = 4;
   m['F'] = 4;
   m['V'] = 4;
   m['5'] = 4;
   m['T'] = 4;
   m['G'] = 4;
   m['B'] = 4;
   m['6'] = 5;
   m['Y'] = 5;
   m['H'] = 5;
   m['N'] = 5;
   m['7'] = 5;
   m['U'] = 5;
   m['J'] = 5;
   m['M'] = 5;
   m['8'] = 6;
   m['I'] = 6;
   m['K'] = 6;
   m[','] = 6;
   m['9'] = 7;
   m['O'] = 7;
   m['L'] = 7;
   m['.'] = 7;
   string s;
   cin >> s;
   for(int i = 0 ; i < s.size() ; i ++){
   	if(!m[s[i]])a[8] ++;
   	else a[m[s[i]]] ++;
   } 
   for(int i = 1 ; i <= 8 ; i ++){
   	cout << a[i] <<endl;
   }
}

T2

当n > m 的时候，每个人都分到一根，再分剩下的
当m > n 的时候，如果m % n == 0每个人能分得n / m ,就是要砍 (m / n - 1 ) * n 刀
否则砍m刀

#include<bits/stdc++.h>
using namespace std;
int n , m;
int ans;
int main () {
    cin >> n >> m;
	while(n % m != 0){
    if(n > m) n %= m;
    int k = m % n == 0 ? m / n - 1 : m / n;
    m = m % n ;
    ans += k * n;
    if(m == 0)break;
	  }
	  cout << ans << endl;
   }

T3

按照右端点排序，左端点先开始并且大于上一个已选活动的结束，就可以加入这个活动

#include<bits/stdc++.h>
using namespace std;
int m ;
struct node{
	int l , r;
}a[10020];
int n ;
bool cmp(node x , node y){
	return x.r < y.r;
}
int b[10020];
int sum[100020];
int main () {
	cin >> m;
	while(m --){
		cin >> n;
		int ans = 0 ;
		for(int i = 1 ; i <= n ; i ++){
			scanf("%d%d" , &a[i].l , &a[i].r);
		}
		sort(a + 1 , a + 1 + n , cmp);
		ans = 1;
		int k = 1; 
		for(int i = 2 ; i <= n ; i ++){
			if(a[i].l >= a[k].r){
				ans ++ ;
				k = i;
			}
		}
		cout << ans << endl;
	}
}

T4

从里到外，显然递归，就遇到左括号开始递归，遇到右括号就返回即可

#include<bits/stdc++.h>
using namespace std;
string calc(){
	int k ;
	string s1 = "" , s2 = "";
	char ch ;
	while(cin >> ch){
		if(ch == '['){
			cin >> k ;
			s2 = calc();
		    while(k --)s1 += s2;
		}
		else if(ch == ']'){
			return s1;
		}
		else{
			s1 += ch;
		}
	}
}
int main () {
	cout << calc() ;
}

T5

按照a * b 排序，最后用高精度即可，高精乘以单精，高精除以单精度

#include<bits/stdc++.h>
using namespace std;
struct node{
	long long l , r;
}p[10020];
bool cmp(node x , node y){
	return x.l * x.r < y.l * y.r;
}
long long n;
long long c1[10200] ;
long long c2[10200] ;
long long c3[10200] ;
void mul(long long x){
	for(int i = 1 ; i <= 10200-1 ; i ++){
		c1[i] *= x;
	}
	for(int i = 1 ; i <= 10200-1  ; i ++){
		c1[i + 1] += c1[i] / 10;
		c1[i] %= 10;
	}
    int l = 10200-1 ;
    while(c1[l] == 0){
    	l --;
	}
/*	for(int i = l ; i >= 1; i --){
		cout << c1[i];
	}  
	cout << endl;*/
}
int len1 = 1 , len2 = 1;
void div(long long x){
    memset(c2 , 0 , sizeof c2);
	int tmp = 0;
	for(int i = 10200-1 ; i >= 1 ; i --){
		tmp *= 10 ;
		tmp += c1[i];
		if(tmp >= x){
			c2[i] = tmp / x;
			tmp %= x;
		}
	}
	len1 = 10200-1;
	while(c2[len1] <= 0 ){
		len1 -- ;
		if(len1 == 1)break;
	}	
}
void fans(){
	if(len1 > len2){
		for(int i = 1 ;i <= len1; i ++){
			c3[i] = c2[i];
		}
		len2 = len1;
	} 
	else if(len1 < len2)return ;
	else if(len1 == len2){
		for(int i = len1 ;i >= 1; i --){
			if(c2[i] > c3[i]){
				for(int j = len1 ;j >= 1;  j --){
					c3[j] = c2[j];
				}
				len2 = len1;
				break;
			}
		}
	}
}
int main () {
	c1[1] = 1;
	cin >> n ;
	cin >> p[0].l >> p[0].r;
	for(int i = 1; i <= n ; i ++){
		cin >> p[i].l >> p[i].r; 
	}
	sort(p + 1 , p + 1 + n , cmp);
	for(int i = 1; i <= n ; i ++){
	  mul(p[i - 1].l);
	  div(p[i].r);
	  fans();
	}
	for(int i = len2 ; i >= 1; i --){
		cout << c3[i];
}
	/*
3
23113 513
213315 5163515
5163516556 15155
5163516556 15155
*/
}