一、题目
Description
Input
Output
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.
二、解题思路
设置体力、感情和智力周期第一次高峰出现时间为A1、A2、A3,假定在X是下次出现三个高峰同天的时间,体力、感情和智力
分别经过了C1、C2、C3个周期。则满足表达式:
X = A1 + C1 * 23;
X = A2 + C2 * 28;
X = A3 + C3 * 33;
根据上式可以推导出:
C1 = (A3 - A1 + C3 * 33) / 23;
C2 = (A3 - A2 + C3 * 33) / 28;
只要针对指定的A1、A2、A3,对于给定的C3根据上式计算有C1和C2都是整数时则就是找的值。
三、代码
#include <iostream>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
struct TimeNode
{
int p;
int e;
int i;
int d;
struct TimeNode* next;
};
int GetNextSameDays(struct TimeNode* pNode)
{
if (pNode==NULL)
{
return -1;
}
int p,e,i;
p = pNode->p % 23;
e = pNode->e % 28;
i = pNode->i % 33;
int nMaxIntell = (21252 + pNode->d) / 33;
int nMinIntell = pNode->d / 33;
int intellectual = nMinIntell;
for (; intellectual <= nMaxIntell ; ++intellectual)
{
/// C1 = (A3 - A1 + C3*33) / 23;
int curIntellectual = i + intellectual*33;
if ( ((curIntellectual - p) % 23) != 0 )
{
continue;
}
if ( ((curIntellectual - e) % 28) != 0)
{
continue;
}
break;
}
if (intellectual > nMaxIntell)
{
return -1;
}
int nSameDays = (intellectual*33 + i - pNode->d);
if (nSameDays <= 0)
{
nSameDays += 21252;
}
return nSameDays;
}
int main()
{
struct TimeNode* pHead = NULL;
struct TimeNode* pCur = NULL;
struct TimeNode* pTemp = NULL;
//pHead = (struct TimeNode*)malloc(sizeof(struct TimeNode));
// pCur = pHead;
int p,e,i,d;
while (cin >> p >> e >> i >> d)
{
if (p ==-1 && e==-1 && i==-1 && d==-1)
{
break;
}
pTemp = (struct TimeNode*)malloc(sizeof(TimeNode));
pTemp->p = p;
pTemp->e = e;
pTemp->i = i;
pTemp->d = d;
pTemp->next = 0;
if (pHead==NULL)
{
pHead = pCur = pTemp;
}
else
{
pCur->next = pTemp;
pCur = pTemp;
}
}
int nCaseIndex = 0;
while (pHead!=NULL)
{
int nNextSameDays = GetNextSameDays(pHead);
if ( nNextSameDays != -1)
{
cout << "Case " << ++nCaseIndex << ":" << " the next triple peak occurs in " << nNextSameDays << " days."<<endl;
}
pTemp = pHead->next;
free(pHead);
pHead = pTemp;
}
return 0;
}