1 solutions

  • 0
    @ 2025-11-5 17:14:14

    C++ :

    // 由于保留5为小数,n > 10000时只要sum = 0.00001即可
    // 只需要循环n 至 n+100 
    
    
    #include<stdio.h>
    
    int main()
    {
    	int kcase = 0;	// 记录数据的组数 
    	int n, m;	
    	while(scanf("%d %d", &n, &m))
    	{
    		if(n == 0 && m == 0)
    		{
    			break;
    		}
    		double sum = 0;
    		if(n >= 10000)
    		{
    			sum = 0.00001;
    			printf("Case %d: %.5lf\n", ++kcase, sum);
    			continue;
    		}
    		int a = n+100 > m ? m : n+100;
    		for(int i = n; i <= a; i++)
    		{
    			sum += 1.0/(i*i);
    		}
    		printf("Case %d: %.5lf\n", ++kcase, sum);
    	}
    	
    	
    	
    	return 0;
    }
    
    • 1

    Information

    ID
    18218
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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