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今天做完 1006題,第一次用枚舉法，但是時間復雜度大，后來看到帖子說是用中國余數法。自己試著也寫了一個，但是用時也過大，現在把代碼貼出來，請大家幫忙改一改啊！謝謝了！
1 #include <stdio.h>
2
3int main(int argc, char* argv[])
4   {
5 int p,e,i,d, index;
6 int day;
7 int x = 28*33*6;
8 int y = 23*33*19;
9 int z = 23*28*2;
10 index = 0;
11
12
13 do
14  {
15 scanf("%d %d %d %d", &p,&e, &i, &d);
16 if(p == -1 && e == -1 && i == -1 && d == -1)
17 break;
18 p = p%23;
19 e = e%28;
20 i = i%33;
21 day = (p*x + e*y + i*z) % 21252;
22 if(day == d)
23 day += 21252;
24 printf("Case %d: the next triple peak occurs in %d days.\n", ++index,day-d >= 0 ? day-d : day-d+21252);
25 }while(1);
26 return 0;
27 }
評論:
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# re: 1006求助
Posted @ 2008-05-31 07:21
/*
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.*/
#include <cstdio>
int main()
{
int x,y,z,s,t;
size_t i(0);
scanf("%d %d %d %d", &x, &y, &z, &s);
while(x!=-1 || y!=-1 || z!=-1 || s!=-1){
t = (5544*x + 14421*y + 1288*z) % 21252;
t -= s;
if(t<=0)
t+=21252;
printf("Case %d: the next triple peak occurs in %d days.\n",
++i,t);
scanf("%d %d %d %d", &x, &y, &z, &s);
}
}
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# re: 1006求助
Posted @ 2008-05-31 07:31
感覺沒多大區別，我的比你少取了點模
40K 75MS
至于哪些超快的，我覺得可能是這個原因：
那句printf應該改成sprintf，先到緩沖區，最后再輸出～
不過就算這樣能提高成績，我覺得其實意義也不大～ 回復 更多評論
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# re: 1006求助
Posted @ 2008-06-01 07:34
有點暈... 回復 更多評論
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# re: 1006求助
Posted @ 2008-06-01 11:34
哦，我在網上看到，有些人寫的程序時間和空間都很少，真不知道是怎么寫出來的 回復 更多評論
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