1.用蠻力解這道題也能AC，雖然測試數據能過但是還是AC了，然后看了一些特殊的數據 24 29 34 2 21251，在修改了一下自己的代碼加了一個else語句，就AC了，可是自己都有點不太明白。暈了?。。。?br>
1
#include <stdio.h>
2#include <stdlib.h>
3
4#define Tp 23
5#define Te 28
6#define Ti 33
7
8int main ()
9
{
10
int p, e, i, d;
11
12 int num = 1;
13
14 while (scanf ("%d%d%d%d", &p, &e, &i, &d) != EOF && (p != -1 && e != -1 && i != -1 && d != -1))
15
{
16 int j;
17
20 for ( j = 1; j <= 21252; j ++)
21 {
22 if ( ((j - p) % Tp == 0) && ((j - e) % Te == 0) && ((j - i) % Ti ==0) )
23 {
24 if ( j - d > 0)
25 printf ("Case %d: the next triple peak occurs in %d days.\n", num, j - d);
26 else
27 printf ("Case %d: the next triple peak occurs in %d days.\n", num, j - d + 21252);
28
29 break;
30
}
31 }
32 num ++;
33 }
34 //system ("pause");
35 return 0;
36
}
37
2.用中國剩余定理解題
顯然下一次高峰出現的時間減去給定的p e i 對周期求余==0；所以利用同余可轉化；
x%Tp = p % Tp = a； x%Te= e % Te = b； x%Ti = i % Tpi= c；
天??！WA了N次居然是因為公式沒理解好 求x時出錯了。
1//2.用中國剩余定理解
2#include <stdio.h>
3#include <stdlib.h>
4int main ()
5{
6 int p, e, i , d, x;
7 int Tp = 23, Te = 28, Ti = 33;
8 int num = 1;
9 while (scanf ("%d%d%d%d", &p, &e, &i, &d ) != EOF && (p != -1 && e != -1 && i != -1 && d != -1) )
10 {
11 int a = p % Tp;
12 int b = e % Te;
13 int c = i % Ti;
14
15 int n1, n2, n3;
16
17 for (int j = 1; j < 33; j ++)
18 {
19 if ( (23 * 28 * j) % 33 == 1)
20 {
21 n1 = j;
22 break;
23 }
24
25 }
26 for (int j = 1; j < 28; j ++)
27 {
28 if ( (23 * 33 * j) % 28 == 1)
29 {
30 n2 = j;
31 break;
32 }
33
34 }
35 for (int j = 1; j < 23; j ++)
36 {
37 if ( (33 * 28 * j) % 23 == 1)
38 {
39 n3 = j;
40 break;
41 }
42
43 }
44
45
46
47 //x = ( n1 * c + n2 * b + n3 * a ) % (23 * 33 * 28);
x = (28 * 23 * n1 * c + 23 * 33 * n2 * b + 28 * 33 * n3 * a ) % (23 * 33 * 28);
48
49 if ( x - d > 0)
50 printf ("Case %d: the next triple peak occurs in %d days.\n", num, x - d);
51 else
52 printf ("Case %d: the next triple peak occurs in %d days.\n", num, x - d + 21252);
53
54 num++;
55 }
56 //system ("pause");
57 return 0;
58}
59
posted on 2010-08-06 15:58
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