输入包括一行,一行中有三个正整数n, t, a(1 ≤ n, t, a ≤ 50), 以空格分割
输出一个整数,表示牛牛可能获得的最高分是多少。
3 1 2
2
#include<iostream>
using namespace std;
int main()
{
int a,n,t;
cin>>n>>t>>a;
cout<<((t>a)?(a+n-t):(t+n-a))<<endl;
return 0;
}
import java.util.*; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub Scanner cin=new Scanner (System.in); int n=cin.nextInt();//总题目数 int t=cin.nextInt();//在考试中牛牛选择√的题目数 int a=cin.nextInt();//n道题目中√的题目数 int max=0; if(t<=a) { max=max+t+n-a; } else { max=max+a+n-t; } System.out.print(max); } }
分三种情况:t=a时,这时的最高分数就是蒙的全部都对;Maxscore=n; t>a时,要保证分数最高也只是蒙的全部都对,对应蒙‘错’的也对 Maxscore=a+n-t; t<a时,要想获得的分数最高蒙的全部都对。Maxscore=t+n-a;#include<iostream> using namespace std; int main() { int a, t, n; int Maxscore=0; cin >> n >> t >> a; if (t == a) Maxscore = n; else if (t > a) Maxscore = a + n - t; else Maxscore = t + n - a; cout << Maxscore << endl; return 0; }
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); int t = sc.nextInt(); //牛牛的正确 int a= sc.nextInt(); //实际上的正确 int noNiu = n - t; //牛牛的错误 int noRel = n - a; //实际上的错误 int sum = 0; if (t > a) { sum += a; } else { sum += t; } if (noNiu > noRel) { sum += noRel; } else { sum += noNiu; } System.out.println(sum); } }
#include<iostream>
using namespace std;
int main() {
int n, t, a;
cin >> n >> t >> a;
if (t == a) {
cout << n << endl;
} else if (t < a) {
int wrong = n - a;
int welldone = wrong + t;
cout << welldone <<endl;
} else {
int wrong = n - a;
wrong = wrong - (t - a);
int welldone = wrong + a;
cout << welldone << endl;
}
return 0;
}
import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; /** * @author wylu */ public class Main { public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String[] strs = br.readLine().split(" "); int n = Integer.parseInt(strs[0]), t = Integer.parseInt(strs[1]), a = Integer.parseInt(strs[2]); System.out.println(Math.min(t, a) + Math.min(n - t, n - a)); } }
#include<iostream> using namespace std; int main(){ int n,t,a; cin>>n>>t>>a; if(t>a) cout<<n-(t-a)<<endl; else cout<<n-(a-t)<<endl; }