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  • hdu 2601 An easy problem

    /*


    此题就是把公式变型,N=i*i +j*j+i*j=(i+1)(j+1)-1,即N+1=(i+1)(j+1)

    即求N+1有几对因数;即需求sqrt(N+1)前面有几个因数:


    */

    #include <stdio.h>
    #include <math.h>
    int main()
    {
    	__int64 n, temp;
    	int i, t, c;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%I64d",&n), n++;  /*n的范围很大*/
    		temp = sqrt((double)n);
    		for(i=2,c=0; i<=temp; i++)
    		{
    			if(n%i == 0) c++;
    		}
    		printf("%d\n",c);
    	}
    	return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/yyf573462811/p/6365363.html
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