smoj—Recaman's Sequence
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这是我前几天就在想的题目,今天总算做出来啦~来写点小总结~嘿嘿~~
这道题的思想很好的,它教我们:先计算出我们需要的数字再从里面挑。这样可以减少程序运行的时间,提高效率。
现在进入正题吧~其实这道题乍一看好像挺简单的,不过这道题也有自己经典的地方哦~
下面是程序:
#include<stdio.h>
#include<string.h>
int a[500002],i,n;
char b[2600000]={0};
void shu()
{
a[0]=0;
b[0]=1;
for(i=1;i<=500000;i++)
{
a[i]=a[i-1]-i;
if(a[i]<0||b[a[i]]==1)
a[i]=a[i-1]+i;
b[a[i]]=1;
}
}
int main()
{
shu();
while(scanf("%d",&n)!=EOF)
{printf("%d\n",a[n]);}
}下午加油了哦~嘿嘿~数字乘积,最大黑区域~come on~~
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