hdu 2058 The sum problem
2012-08-03 18:24
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//这种题用暴力肯定会超时,题目的突破点就是相邻的值相差值是相等的,利用等差数列公式sum=a1*n+(n-1)*n/2
//知道到a1=1时,n是最大的,也就是sqrt(2*m),从最大的开始,每次减1,来求a1的值,a1=(sum/n)-(n-1)/2, 在把a1和n代进去,看是否相等
#include"stdio.h"
#include"math.h"
int main()
{
int n,m,a1,nn;
while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
{
nn=(int)sqrt(2.0*m);
while(nn)
{
a1=m/nn-(nn-1)/2;
if(a1*nn+(nn-1)*nn/2==m)
printf("[%d,%d]\n",a1,nn+a1-1);
nn--;
}
printf("\n");
}
return 0;
}
//这种题用暴力肯定会超时,题目的突破点就是相邻的值相差值是相等的,利用等差数列公式sum=a1*n+(n-1)*n/2
//知道到a1=1时,n是最大的,也就是sqrt(2*m),从最大的开始,每次减1,来求a1的值,a1=(sum/n)-(n-1)/2, 在把a1和n代进去,看是否相等
#include"stdio.h"
#include"math.h"
int main()
{
int n,m,a1,nn;
while(scanf("%d%d",&n,&m)!=EOF&&(n||m))
{
nn=(int)sqrt(2.0*m);
while(nn)
{
a1=m/nn-(nn-1)/2;
if(a1*nn+(nn-1)*nn/2==m)
printf("[%d,%d]\n",a1,nn+a1-1);
nn--;
}
printf("\n");
}
return 0;
}
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