学步园

题意：有3个循环周期，周期天数分别为23、28、33。对于某一年，已知某年这3个周期的某一峰值分别是当年的第p、e、i天，问从第d天开始到最近一个满足3个周期都达到峰值的日期还有多少天。

数据比较小，直接打表可过：

# include <stdio.h>
int tab[40][40][40]; 
int cal(int num)
{
	if (num <= 0) return num + 21252 ;
	return num ;
}

int main ()
{
	int p,e,i,d,T, nCase ;
	for (i = 0 ; i < 21252 ; i++)
		tab[i%23][i%28][i%33] = i ; 
	scanf ("%d", &T) ;
	while (T--)
	{
		nCase = 1 ;
		while (1)
		{
			scanf ("%d%d%d%d", &p, &e, &i, &d) ;
			if (p==-1&&e==-1&&i==-1&&d==-1) break ;
			printf ("Case %d: "
				"the next triple peak occurs in %d days.\n",
				nCase++, cal(tab[p%23][e%28][i%33]-d)) ;
		}
		if (T) putchar ('\n') ;
	}
	return 0 ;
}

中国剩余定理效率更高：0MS

#include <iostream>
using namespace std;
int exgcd(int a,int b,int &x,int &y)
{
	if(a==0)
	{
		x=0;
		y=1;
		return b;
	}
	int g = exgcd(b%a,a,x,y);
	int tem = y;
	y=x;
	x=tem-(b/a)*y;
	return g;
}
int inv(int a,int n)//求逆元
{
	int x,y;
	exgcd(a,n,x,y);
	return (x%n+n)%n;
}
int main()
{
	int N,P,E,I,D,ans;
	const int m1=23,m2=28,m3=33,M1=28*33,M2=23*33,M3=23*28,m=23*28*33;
	const int M11 = inv(M1,m1),M22 = inv(M2,m2),M33 = inv(M3,m3);//求mi的逆元 
	scanf("%d",&N);
	while(scanf("%d%d%d%d",&P,&E,&I,&D)!=EOF)
	{
		if(P==-1&&E==-1&&I==-1&&D==-1)break;
		//a[i]*(mi*mi(逆)mod ni)因为在求逆元的时候已经mod ni了 
		//(mi*mi(逆)mod ni)是ci 
		ans = (P*M1*M11 + E*M2*M22 + I*M3*M33)%m;
		ans -= D;
		if(ans<=0) ans+=m;
		printf("Case %d: the next triple peak occurs in %d days.\n",N++,ans);
		//cout<<"Case "<<N++<<": the next triple peak occurs in "<<i<<" days.\n";
	}
	return 0;
}