POJ1006 Biorhythms【中国剩余定理】

题目:

Description

Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical, emotional, and intellectual cycles, and they have periods of lengths 23, 28, and 33 days, respectively. There is one peak in each period of a cycle. At the peak of a cycle, a person performs at his or her best in the corresponding field (physical, emotional or mental). For example, if it is the mental curve, thought processes will be sharper and concentration will be easier. 
Since the three cycles have different periods, the peaks of the three cycles generally occur at different times. We would like to determine when a triple peak occurs (the peaks of all three cycles occur in the same day) for any person. For each cycle, you will be given the number of days from the beginning of the current year at which one of its peaks (not necessarily the first) occurs. You will also be given a date expressed as the number of days from the beginning of the current year. You task is to determine the number of days from the given date to the next triple peak. The given date is not counted. For example, if the given date is 10 and the next triple peak occurs on day 12, the answer is 2, not 3. If a triple peak occurs on the given date, you should give the number of days to the next occurrence of a triple peak. 

Input

You will be given a number of cases. The input for each case consists of one line of four integers p, e, i, and d. The values p, e, and i are the number of days from the beginning of the current year at which the physical, emotional, and intellectual cycles peak, respectively. The value d is the given date and may be smaller than any of p, e, or i. All values are non-negative and at most 365, and you may assume that a triple peak will occur within 21252 days of the given date. The end of input is indicated by a line in which p = e = i = d = -1.

Output

For each test case, print the case number followed by a message indicating the number of days to the next triple peak, in the form: 
Case 1: the next triple peak occurs in 1234 days. 
Use the plural form ``days'' even if the answer is 1.

Sample Input

0 0 0 0
0 0 0 100
5 20 34 325
4 5 6 7
283 102 23 320
203 301 203 40
-1 -1 -1 -1

Sample Output

Case 1: the next triple peak occurs in 21252 days.
Case 2: the next triple peak occurs in 21152 days.
Case 3: the next triple peak occurs in 19575 days.
Case 4: the next triple peak occurs in 16994 days.
Case 5: the next triple peak occurs in 8910 days.
Case 6: the next triple peak occurs in 10789 days.
思路:

这道题很简单。就是三个周期为23,28,33,在输入的第a天,第b天,第c天分别同时达到周期的顶峰,求下次同时达到顶峰的日子与d的相差天数。

需要注意的是,输入的a,b,c不一定为从这一年起的第一个周期!见题目我标红的字。

当时忽视了这句话,WA了好多次,哭哭啊!

当时错的测试用例:1 1 1 0, 答案应该为1,而我答案为21253。这个好解决,把i=0的情况加上去就好了。

后来想到,有一个测试用例我当时应该也错了:1 1 34 0,因为我每次都是从c开始加,而c不一定是这一年起第一个33的周期啊!

所以后来我用

while(c-33>0)
	c-=33;
算出了c的第一个顶峰。

终于AC了。结果是125ms,好像还是不太快啊。
代码:

#include<iostream>
using namespace std;
int main(){
	int a,b,c,d;
	int count=0;
	while(cin>>a){
		if(a==-1)
			break;
		count++;
		cin>>b>>c>>d;
		bool isFind=false;
		int i=0;
		int newDay=0;
		while(isFind==false){
			while(c-33>0)
				c-=33;
			newDay=c+i*33;
			if((newDay-b)%28==0&&(newDay-a)%23==0)
				isFind=true;
			i++;
		}
		if(newDay<=d)
			newDay+=21252;
		cout<<"Case "<<count<<": the next triple peak occurs in "<<(newDay-d)<<" days."<<endl;
	}
	return 0;
}

后来去看别人的代码,他们用到了中国剩余定理,然后直接套公式就做出来,时间可以达到0ms。

看不懂中国剩余定理为啥公式是那样子,只好死记硬背了QAQ。

再看我们这道题,读入p,e,i,d 4个整数,求n。
已知(n+d)%23=p; (n+d)%28=e; (n+d)%33=i ,求n 。

两道题是一样的。但是韩信当时计算出结果的?
韩信用的就是“中国剩余定理”,《孙子算经》中早有计算方法,大家可以查阅相关资料。
“韩信点兵”问题计算如下:
因为n%3=2, n%5=3, n%7=2 且 3,5,7互质 (互质可以直接得到这三个数的最小公倍数)
令x= n%3=2 , y= n%5=3 ,z= n%7=2
使5×7×a被3除余1,有35×2=70,即a=2,用70;
使3×7×b被5除余1,用21×1=21,即b=1,用21;
使3×5×c被7除余1,用15×1=15,即c=1,用15。
那么n =(70×x+21×y+15×z)%lcm(3,5,7) = 23 这是n的最小解(Icm表示最小公倍数)
而韩信已知士兵人数在2300~2400之间,所以只需要n+i×lcm(3,5,7)就得到了2333,此时i=22

同样,这道题的解法就是:
已知(n+d)%23=p; (n+d)%28=e; (n+d)%33=i
使33×28×a被23除余1,用33×28×8=5544;
使23×33×b被28除余1,用23×33×19=14421;
使23×28×c被33除余1,用23×28×2=1288。
因此有(5544×p+14421×e+1288×i)% lcm(23,28,33) =n+d
又23、28、33互质,即lcm(23,28,33)= 21252;
所以有n=(5544×p+14421×e+1288×i-d)%21252
本题所求的是最小整数解,避免n为负,因此最后结果为n= [n+21252]% 21252
那么最终求解n的表达式就是:
n=(5544*p+14421*e+1288*i-d+21252)%21252;
当问题被转化为一条数学式子时,你会发现它无比简单。。。。直接输出结果了。

下面是代码:

#include<iostream>
using namespace std;

int main(void)
{
	int p,e,i,d;
	int time=1;
	while(cin>>p>>e>>i>>d)
	{
		if(p==-1 && e==-1 && i==-1 && d==-1)
			break;

		int lcm=21252;  // lcm(23,28,33)
		int n=(5544*p+14421*e+1288*i-d+21252)%21252;
		if(n==0)
			n=21252;
		cout<<"Case "<<time++<<": the next triple peak occurs in "<<n<<" days."<<endl;
	}
	return 0;
}

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