1621: [Usaco2008 Open]Roads Around The Farm分岔路口
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.
模拟即可。。。。
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 using namespace std; 7 long long ans=0; 8 void dif(long long x,long long y) 9 { 10 int k=(x-y)%2; 11 if(k==0&&x>1&&x>y) 12 { 13 dif((x-y)/2,y); 14 dif((x-y)/2+y,y); 15 } 16 else 17 { 18 ans++; 19 return ; 20 } 21 } 22 int main() 23 { 24 long long n,k; 25 scanf("%lld %lld",&n,&k); 26 dif(n,k); 27 printf("%lld",ans); 28 }