1 solutions

  • 0
    @ 2025-11-5 15:52:08

    C :

    #include<stdio.h>
    #include<string.h>
    int main()
    {
    	char x[10][25];
    	int i,j,l[10],y[10],n;
    	scanf("%d",&n);
    	for(i=0;i<n;i++)
    	{
    		getchar();
    		scanf("%s %d",x[i],&y[i]);
    		l[i]=strlen(x[i]);
    	}
    	for(i=0;i<n;i++)
    	{
    		if(y[i]>l[i]-2)	printf("Error");
    		else	printf("%c",x[i][y[i]+1]);
    		if(i!=n-1)
    		printf("\n");
    	}
    }
    

    C++ :

    #include<bits/stdc++.h>
    using namespace std;
    int n;
    string a;
    double s;
    int wz,ss,flag,y;
    int main()
    {
    	cin>>n;
    	getchar();
    	for(int i=1;i<=n;i++)
    	{
    		cin>>a;
    		cin>>s;
    		for(int j=0;j<a.size();j++)
    		{
    			if(a[j]=='.')
    				flag=1;
    			if(flag==0)y++;
    		}
    		if(y+1+s<=a.size())cout<<a[s+1]<<endl;
    		else cout<<"Error"<<endl;
    		flag=0;
    		ss=0;
    		y=0;
    	}
       return 0;
    }
    
    
    • 1

    【创新型】第10章: 字符串 10.25 调皮的小数点

    Information

    ID
    17271
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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