BZOJ 1621: [Usaco2008 Open]Roads Around The Farm分岔路口
2017-08-27 16:21
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1621: [Usaco2008 Open]Roads Around The Farm分岔路口
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 891 Solved: 660
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Description
约翰的N(1≤N≤1,000,000,000)只奶牛要出发去探索牧场四周的土地.她们将沿着一条路走,一直走到三岔路口(可以认为所有的路口都是这样的).这时候,这一群奶牛可能会分成两群,分别沿着接下来的两条路继续走.如果她们再次走到三岔路口,那么仍有可能继续分裂成两群继续走. 奶牛的分裂方式十分古怪:如果这一群奶牛可以精确地分成两部分,这两部分的牛数恰好相差K(1≤K≤1000),那么在三岔路口牛群就会分裂.否则,牛群不会分裂,她们都将在这里待下去,平静地吃草. 请计算,最终将会有多少群奶牛在平静地吃草.Input
两个整数N和K.Output
最后的牛群数.Sample Input
6 2INPUT DETAILS:
There are 6 cows and the difference in group sizes is 2.
Sample Output
3OUTPUT DETAILS:
There are 3 final groups (with 2, 1, and 3 cows in them).
6
/ \
2 4
/ \
1 3
HINT
6只奶牛先分成2只和4只.4只奶牛又分成1只和3只.最后有三群奶牛.Source
Silver题解:同样是水题啊(别D我)。。。DFS就过了。。。
代码:
#include<cstdio>
using namespace std;
int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,k,ans;
void dfs(int n,int k)
{
if(n<2+k||(n+k)%2==1)
{
ans++;
return;
}
dfs((n+k)/2,k);
dfs((n-k)/2,k);
}
int main()
{
n=read();k=read();
dfs(n,k);
printf("%d\n",ans);
return 0;
}
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