1 solutions

  • 0
    @ 2025-11-5 18:01:04

    C :

    #include<stdio.h>
    #include<math.h>
    int main()
    {
    	int y,i;
    	double sum;
    	while(scanf("%d",&y)!=EOF,y)
    	{
    		sum=0;
    		if(y==1960)
    			printf("3\n");
    		else
    		{
    			for(i=1;;i++)
    			{
    				sum+=log10(i)/log10(2);
    				if(sum>pow(2,(y-1960)/10+2))
    					break;
    			}
    			printf("%d\n",i-1);
    		}
    	}
    	return 0;
    }
    
    • 1

    Information

    ID
    18715
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
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