UVa 10940 - Throwing cards away II(数论,规律)
2013-07-24 19:29
309 查看
这个题可以按逆序思路来想
最后剩下的牌,除了1张之外,最后一张剩下的一定是偶数牌
然后,2^n张牌剩下的一定是2^n
2^n-1张牌剩下的一定是2^n-2
...
以此,可推得以下数列:
ans(1)=1;
ans(2)=2;
ans(3...4)={2,4};
ans(5...8)={2,4,6,8};
...
ans(2^(n-1)+1...2^n)={2,...,2^n}
#include <stdio.h>
int main()
{
int p[500010];
int s=3,n;
p[1]=1;
p[2]=2;
int k=1;
while(1)
{
if(s>500000)break;
for(int j=1; j<=(1<<k); j++)
{
if(s>500000)break;
p[s++]=j*2;
}
k++;
}
while(scanf("%d",&n)&&n)
{
printf("%d\n",p
);
}
return 0;
}
最后剩下的牌,除了1张之外,最后一张剩下的一定是偶数牌
然后,2^n张牌剩下的一定是2^n
2^n-1张牌剩下的一定是2^n-2
...
以此,可推得以下数列:
ans(1)=1;
ans(2)=2;
ans(3...4)={2,4};
ans(5...8)={2,4,6,8};
...
ans(2^(n-1)+1...2^n)={2,...,2^n}
#include <stdio.h>
int main()
{
int p[500010];
int s=3,n;
p[1]=1;
p[2]=2;
int k=1;
while(1)
{
if(s>500000)break;
for(int j=1; j<=(1<<k); j++)
{
if(s>500000)break;
p[s++]=j*2;
}
k++;
}
while(scanf("%d",&n)&&n)
{
printf("%d\n",p
);
}
return 0;
}
相关文章推荐
- UVA 10940 - Throwing cards away II(规律)
- UVA10940 - Throwing cards away II(找规律)
- UVA 10940 Throwing cards away II
- uva 10940 - Throwing cards away II
- UVa 10940 - Throwing cards away II
- UVA 10940 Throwing cards away II
- uva 10940 Throwing cards away II
- uva 10940 Throwing cards away II
- UVa 10940 Throwing cards away II (约瑟夫问题)
- UVA10940 - Throwing cards away II(找到规律)
- Uva10935 - Throwing cards away I
- Throwing cards away II
- UVA 10935 - Throwing cards away I
- uva10935 - Throwing cards away I
- UVA10935 Throwing cards away I【模拟+queue+循环队列】
- Throwing cards away I UVA - 10935
- UVa 10935 - Throwing cards away I
- UVA 10935 - Throwing cards away I
- (1.1.7)UVA 10935 Throwing cards away I(直叙式模拟)
- UVa 10935 Throwing cards away I