速通51题解

‮队团加不）ด้้童帅_者仇复

2024-05-08 13:18:47

发布于：广东

29阅读

0回复

0点赞

T1 A + B + C
T2 2的N次方
 T3 下一个字母
 T4 千位分割符
 T5 画个ACGO
T6 音乐播放器
 T7 10的N次方
 T8 日期间隔计算
 T9 ZXC的闯关计划
 T10 2024的倍数
代码摆这了，要讲解看题解（除T1太简单）

T1

#include <iostream>
#include <cstdio>
using namespace std;
int a[100005];
int main(){
	int a, b, c;
	cin >> a >> b >> c; 
	cout << a + b + c;

	return 0;
}

T2

#include <iostream>
#include <cstdio>
using namespace std;
int a[100005];
int main(){
	long long x = 1;
	int n;
	cin >> n;
	cout << (x << n);

	return 0;
}

T3

#include <iostream>
#include <cstdio>
using namespace std;
int a[100005];
int main(){
	string a;
	cin >> a;
	for(int i = 0; a[i] != '\0'; i++){
		if('a' <= a[i] && a[i] <= 'z'){
			if(a[i] == 'z') a[i] = 'a';
			else a[i]++;
		}else if(a[i] >= 'A'){
			if(a[i] == 'Z') a[i] = 'A';
			else a[i]++;
		}
	}cout << a;

	return 0;
}

T4

#include <iostream>
#include <cstdio>
using namespace std;
int a[100005];
void fenge(int n){
	if(n < 1000){
		cout << n;
		return;
	}fenge(n / 1000);
	printf(",%03d", n % 1000);
}
int main(){
	int n;
	cin >> n;
	fenge(n);

	return 0;
}

T5

#include <iostream>
#include <cstdio>
using namespace std;
int a[100005];
int main(){
	cout << R"(     ___       ______   _______   ______   
    /   \     /      | /  _____| /  __  \  
   /  ^  \   |  ,----'|  |  __  |  |  |  | 
  /  /_\  \  |  |     |  | |_ | |  |  |  | 
 /  _____  \ |  `----.|  |__| | |  `--'  | 
/__/     \__\ \______| \______|  \______/)";

	return 0;
}

T6

#include <iostream>
#include <cstdio>
using namespace std;
int main(){
	int n, m, k;
	cin >> n >> m >> k;
	string a;
	while(m--){
		cin >> a;
		if(a == "<-") k--;
		else k++;
	}
	if(k % n == 0) cout << n;
	else cout << k % n;

	return 0;
}

T7

#include <iostream>
#include <cstdio>
using namespace std;
int a[100005];
int main(){
	int n;
	cin >> n;
	cout << 1;
	while(n--) cout << 0;

	return 0;
}

T8

#include <iostream>
#include <cstdio>
using namespace std;
int mth[] = {0, 31, 0, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n1, m1, k1, n2, m2, k2;
bool check(int x){
	return x % 400 == 0 || x % 4 == 0 && x % 100;
}void s(){
	swap(n1, n2), swap(m1, m2), swap(k1, k2);
}
int main(){
	scanf("%d-%d-%d %d-%d-%d", &n1, &m1, &k1, &n2, &m2, &k2);
	if(n1 > n2){
		s();
	}else if(n1 == n2 && m1 > m2){
		s();
	}else if(n1 == n2 && m1 == m2 && k1 > k2){
		s();
	}int ct = 0;
	while(n1 != n2 || m1 != m2 || k1 != k2){
		if(check(n1)) mth[2] = 29;
		else mth[2] = 28;
		k1++, ct++;
		if(k1 > mth[m1]) m1++, k1 = 1;
		if(m1 >= 13) n1++, m1 = 1;
	}cout << ct;

	return 0;
}

T9

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[100005];
int main(){
	long long n, m;
	cin >> n >> m;
	for(int i = 1; i <= n; i++){
		cin >> a[i];
	}sort(a + 1, a + n + 1);
	for(int i = 1; i <= n; i++){
		if(m <= a[i]){
			cout << -1;
			return 0;
		}m -= a[i];
		m *= 1.02;
	}
	cout << m;

	return 0;
}

T10

#include <iostream>
#include <cstdio>
using namespace std;
long long bucket[100005];
int main(){
	int n, x;
	cin >> n;
	for(int i = 1; i <= n; i++){
		cin >> x;
		bucket[x % 2024]++;
	}long long ct = 0;
	for(int i = 0; i <= 2023; i++){
		ct += bucket[i] * (bucket[i] - 1) / 2;
	}cout << ct;

	return 0;
}