1 条题解

  • 0
    @ 2025-4-7 21:19:29

    C :

    #include<stdio.h>
    int main()
    {
    	int Xcm(int a,int b);
    	int n,m,i,x,a[20],s;//m为最高次数幂,数组a[20]存下每次输入的数
    	scanf("%d",&n);//i为计时器,s计算每个表达式的值
    	while(n--)
    	{
    		scanf("%d",&m);
    		for(i=0;i<=m;++i)
    			scanf("%d",&a[i]);
    		scanf("%d",&x);
    		s=a[0];
    		for(i=1;i<=m;++i)
    			s+=a[i]*Xcm(x,i);
    		printf("%d\n",s);
    	}
    	return 0;
    }
    int Xcm(int a,int b)
    {
    	int i,s=a;
    	for(i=1;i<b;++i)
    		s=s*a;
    	return s;
    }
    

    C++ :

    #include <iostream>
    #include <cstdio>
    using namespace std;
    
    int main()
    {
        //freopen("test.in", "r", stdin);
        //freopen("test.out", "w", stdout);
        int n, x, t;
        int num[20];
        cin >> t;
        while (t--) {
            cin >> n;
            for (int i = 0; i <= n; ++i)
                cin >> num[i];
            cin >> x;
            int s = 0;
            for (int i = n; i >= 0; --i)
                s = s * x + num[i];
            cout << s << endl;
        }
        return 0;
    }
    
    

    Java :

    
    
    import java.util.Scanner;
    
    public class Main{
       private static Scanner s = new Scanner(System.in) ;
       
       public static void main(String[] args) {
    	 int t = s.nextInt() ;
    	 for (int j = 0; j < t; j++) {
    		 int n = s.nextInt() ;
    		 int a[] = new int[n+1] ;
    		 for (int i = 0; i <= n; i++) {
    			a[i] = s.nextInt() ;
    		 }
    		 int x = s.nextInt() ;
    		 int sum = 0 ;
    		 int c = 1 ;
    		 for (int i = 0; i < a.length; i++) {
    				
    			sum = sum+ c*a[i];
    			c = x*c ;
    		 }
    		 
    		 System.out.println(sum);
    	    }
    	}
    	 
    }
    
    
    • 1

    信息

    ID
    1157
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者