UVa 12531 - Hours and Minutes
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題目:給你一個時鐘的角度,問時鐘能形成這個角度。
分析:數論,群論。判斷是否能整除6即可,下面證明。
必要性:分針每次移動為360/60 = 6度,時針每次移動為30/60 = 0.5度;
這裡分針必須每次移動12度,才能滿足角度差為整度數;
而此時,時針和分針都指向60個分度之一,所以一定差6的倍數度;
充分性:這裡比較重要,要證明所有的6的倍數的角度都能取到;
根據上面分析,分針的運動角度為12k,而時針的運動角度為k;
則夾角為(12k%360+360-k)%360,简化式子如下;
(12k%360+360-k)%306= 12k%360+(360-k)%360= 11k%360;
這裡利用性质:gcd(a,b)= 1,则ak%b构成的前b个数字不重复;
(反证法,假设i≠j,且ai%b = aj%b,则a(i-j)%b = 0;
又因為a,b互質;则i-j是b的倍数,這與 0 < i,j < b矛盾)
11k%360构成的数字中前360个不会重复,而12小时正好有360组数据。
#include <cstring>#include <cstdio>int main(){int n;while (~scanf("%d",&n))if (n%6 == 0)puts("Y");else puts("N"); return 0;}
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