1 条题解

  • 0
    @ 2024-12-26 17:11:20

    C :

    #include<stdio.h>
    int main() 
    {
    	int t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		int x,y,k,ans;
    		scanf("%d%d%d",&x,&y,&k);
    		if(y<=x)
    		ans=x;
    		else if(k>=y-x)
    		ans=y;
    		else
    		ans=y+y-x-k;
    		printf("%d\n",ans);
    	}
    	return 0;
    } 
    

    C++ :

    #include <bits/stdc++.h>
    #define int long long
    using namespace std;
    const int N=1e6+10;
    typedef pair<int,int> PII;
    typedef priority_queue<int , vector<int>, greater<int>> minqueue;  //从小到大 queue
    typedef priority_queue<int, vector<int>, less<int>> maxqueue;      //从大到小 queue
    
    int gcd(int a, int b){  //最大公因数 
        return b ? gcd(b, a % b) : a;
    }
    
    int lcm(int a, int b){ //最小公倍数 
        return a * b / __gcd(a, b);
    }
    
    int qmi(int base, int power, int p)  //快速幂求余 
    {
    	int result = 1;   
    	while (power > 0)           
    	{
    		if (power & 1)         							
    			result = result * base % p;   
    		base = base * base % p ;       						
    		power >>= 1;         						
    	}
    	return result % p;       
    }
    
    void solve()
    {
    	int x,y,k;
    	cin>>x>>y>>k;
    	if(x>=y){
    		cout<<x<<'\n';
    	}
    	else{
    		if(x+k>=y) cout<<y<<'\n';
    		else cout<<y+y-(x+k)<<'\n';
    	}
    }
    
    signed main()
    {
    	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    	int t;
    	t=1;
    	cin>>t; 
    	while(t--)     
    	{
    		solve();
    	}
    	return 0;
    }
    

    Java :

    import java.util.Scanner;
    
    public class Main {
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int t= sc.nextInt();
        for (int i = 0; i <t; i++) {
            int x= sc.nextInt();
            int y= sc.nextInt();
            int k= sc.nextInt();
            int r;
            if(x>y){
                r=x;
                System.out.println(r);
            }
             else if(x<y&&(x+k)>=y){
                r=y;
                System.out.println(r);
            } else if (x<y&&(x+k)<y) {
                r=2*y-x-k;
                System.out.println(r);
            }
    
        }
    }
    }
    
    
    • 1

    信息

    ID
    722
    时间
    1000ms
    内存
    128MiB
    难度
    10
    标签
    提交数
    8
    已通过
    2
    上传者