2022 SMU-XCPC 集训队新生赛

# 游乐园

因为套票更加便宜，根据贪心应该尽可能的选择套票。

#include<bits/stdc++.h>
using namespace std;

int main(){
    int x , y;
    cin >> x >> y;
    cout << 60 * x + 40 * y - 10 * min( x , y );
}

口算练习

认真读读题就好了

#include<bits/stdc++.h>
using namespace std;
 
int32_t main() {
    int a , b;
	cin >> a >> b;
    cout << (int) floor( 1.0 * a / b ) << "\n";
    return 0;
}

群管理员

循环判断就好，但是要注意数据范围

#include<bits/stdc++.h>
#define int long long
using namespace std;
 
int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}
 
int32_t main() {
    int n = read() , k = read() , cnt = 0;
    for( int x ; n ; n -- )
        x = read() , cnt += ( k == x );
    cout << cnt << "\n";
    return 0;
}

集卡

依旧是循环判断就好，但是注意空间限制不能使用数组

#include<bits/stdc++.h>

using namespace std;

int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    for( int t = read() , n , x , flag ; t ; t -- ){
        n = read() , flag = 0;
        for(  ; n ; n -- ){
            x = read();
            if( x == 0 ) flag = 1;
        }
        cout << ( flag ? "yes\n" : "no\n" );
    }
    return 0;
}

排队

简单的分一下奇偶数就好了

#include<bits/stdc++.h>
#define int long long
using namespace std;


int read() {
    int x = 0, f = 1, ch = getchar();
    while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
    if (ch == '-') f = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
}

int32_t main() {
    vector<int> a , b;
    for( int n = read() , x ; n ; n -- ){
        x = read();
        if( x & 1 ) a.push_back(x);
        else b.push_back(x);
    }
    for( int it : a )
        printf("%d " , it );
    printf("\n");
    for( int it : b )
        printf("%d " , it );
    printf("\n");
    return 0;
}

smuxcpc

直接遍历一下字符串就好了

#include<bits/stdc++.h>
using namespace std;

int main(){
    string s;
    int ans=0;
    cin>>s;
    for(int i=0;i<s.length();i++){
        if(s[i]=='s'){
       if(s[i+1]=='m'&&s[i+2]=='u'&&s[i+3]=='x'&&s[i+4]=='c'&&s[i+5]=='p'&&s[i+6]=='c')
                ans++;
        }

    }
    cout<<ans<<endl;
    return 0;
}

可以直接用python 写

print(input().count("smuxcpc"))

也可用用新语法

#include <bits/stdc++.h>

#define ll long long
#define ull unsigned long long
#define endl (char)10

using namespace std;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);cout.tie(nullptr);

	int ans = 0, op = 0;
	string s; cin >> s;

	for (int tmp = s.find("smuxcpc", op), op = tmp + 1; tmp != s.npos; tmp = s.find("smuxcpc", op), op = tmp + 1) {
		ans++;
	}
	cout << ans << endl;
	return 0;
}

军训

这题其实算是本厂最困难的一道题了，我们要学会读题，使用题目给出的数据生成器生成序列和询问区间，注意对于最小值应该输出编号

#include<bits/stdc++.h>
using namespace std;


int data( int & seed , int n ){
    seed = (seed % 123456 + seed % 10000007) % n + 1;
    return seed;
}

int32_t main() {
    int n , m ,seed;
    cin >> n >> m >> seed;
    vector<int> a(n+1);
    for( int i = 1 ; i <= n ; i ++ )
        a[i] = data( seed , n );


    for( int i = 1 ,op , l , r ; i <= m ; i ++ ){
        op = data( seed , n ) , l = data( seed , n ) , r = data( seed , n );
        if( l > r ) swap( l , r );
        if( op & 1 ){
            int res = INT_MIN;
            for( int i = l ; i <= r ; i ++ )
                res = max( res , a[i] );
            cout << res << "\n";
        }else{
            int res = INT_MAX , id = 0;
            for( int i = l ; i <= r ; i ++ )
                if( a[i] < res ) res = a[i] , id = i;
            cout << id << "\n";
        }
    }
    return 0;
}

omemi的心事

这道题应该才是最简单一题，注意题目并不是难度递增的哟

#include <iostream>
using namespace std;

int main(){
    char c;
    cin >> c;
    cout << (int)c;
}