BZOJ1621: [Usaco2008 Open]Roads Around The Farm分岔路口
1621: [Usaco2008 Open]Roads Around The Farm分岔路口
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 521 Solved: 380
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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 2
INPUT DETAILS:
There are 6 cows and the difference in group sizes is 2.
INPUT DETAILS:
There are 6 cows and the difference in group sizes is 2.
Sample Output
3
OUTPUT DETAILS:
There are 3 final groups (with 2, 1, and 3 cows in them).
6
/ \
2 4
/ \
1 3
OUTPUT 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
题解:
模拟一下即可
代码:
View Code
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set> 10 #include<queue> 11 #define inf 1000000000 12 #define maxn 300+10 13 #define maxm 500+100 14 #define ll long long 15 using namespace std; 16 inline ll read() 17 { 18 ll x=0,f=1;char ch=getchar(); 19 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 20 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 21 return x*f; 22 } 23 int n,k,ans=0; 24 void dfs(int x) 25 { 26 if(x-k<2||((x&1)!=(k&1)))ans++; 27 else 28 { 29 int y=(x-k)>>1;dfs(y);y=x-y;dfs(y); 30 } 31 } 32 int main() 33 { 34 freopen("input.txt","r",stdin); 35 freopen("output.txt","w",stdout); 36 n=read();k=read(); 37 dfs(n); 38 printf("%d\n",ans); 39 return 0; 40 }